Well, that was a very brute-force approach to this problem. I feel like we should let a computer do this kind of repetitive work if we're going to go through this much trouble. I wonder what elegant methods may get us here? Some were mentioned early knocking out possibilities.
Yeah... I was expecting some kind of cool solution myself, but this was pretty much brute force. Is like showing us a maze, and asking "How could we solve this maze?" and then it starts drawing all the paths to see if it reaches the end of the maze at some point. XD
I think the computation could have been reduced pretty quickly by noting how big the product could be, considering you have to add a 2 digit number to that and still get a 2 digit number.
So, my mental reasoning for this one went: - Final result must be under 100 - We don't have a zero - Number after multiplication must leave us digits that can add to 98 or below You can eliminate the vast majority of options from this, so what next... - Multiplier cannot be 1; it would cause repeat digits. - Multiplier cannot be five; it would cause repeat digits. - Multiplier cannot be above five, as this doesn't leave any way of leaving you a second two-digit number that will keep you under 100. - ((Multiplier must be 2, 3 or 4)) - Second digit of initial value cannot be 5; it would cause repeats, or a zero. - Second digit of initial value cannot be 1; any result would require another 1 and be invalid. - First digit in product cannot be 1; this is impossible mathematically (we are multiplying at least a two digit number by at least 2). - First digit in the final answer also cannot be 1, for the same reason. - Second digit in product could only by 1 if the multiplier is 3 AND second digit in initial value is 7; no other combination allows this. (at this point I'd mentally eliminated a lot, but changed tack slightly without an immediate path of eliminating to a definite position for the 1) - If the multiplier must be 2, 3 or 4, we can begin checking some combinations, because we'll quickly be able to eliminate whole branches of inquiry without needing to check. - If we start with a multiplier of 4, we can only possibly have to check initial values in the 1(?) and 2(?) range - anything higher prevents us staying 98 or under. - Of the x4 multiplier options, we can eliminate any combinations where the second digit is 4. - Of the x4 multiplier options with an initial value of 2(?), we eliminate 21, 22, 24 and anything over 24 (product over 98) automatically. 23 can also be eliminated (4 x 3 = 12, 2 is used) - Of the x4 multiplier options with an initial value of 1(?), 12, 14, 15 and 16, are eliminated automatically for repeat digits, leaving just 13, 17, 18 and 19. - 13, 18 and 19 all leave us with remaining digits for which even the smallest possible addition will take us over 98. (52 with [6, 7, 8, 9] left, 72 with [3, 5, 6, 9] left and 76 with [2, 3, 5, 8] left) - I couldn't think of an immediate way to eliminate 17, so I checked it manually. - 17 x 4 = 68 [2, 3, 5, 9] Lead digit could only be 2, as anything else goes over 98. Second digit could only be 5, as anything else leaves an invalid unit digit. - 17 x 4 = 68 + 25 = 93 - Success! 1 Valid Solution, then, seems to be: 17 x 4 = 68 + 25 = 93 == I've been told there's only one solution, and this solution is valid, but I've not *proven* that there is only one solution; my tester might be mistaken, so I continued. We've eliminated all other x4 multipliers, so the only place where another solution could exist would be in the x2 and x3 sets. Still working down, I started thinking about what else I could eliminate from the x3 set. - Of the x3 multiplier options, we can eliminate all initial values over 2(?) as well, just like for x4 multipliers; we can't use 3(?) for repeat digits. - If we start with a multiplier of x3, we can only possibly have to check initial values in the 1(?) and 2(?) range; we can't use 3(?), and 4(?) puts us over. - Of the x3 multiplier options, we eliminate any combination where the second digit is 3. - Of the x3 multiplier options with an initial value of 2(?), we've already eliminated 21, 22, 23 and 25, and can quickly eliminate 24 and 28 (3 x 4 = 12, repeat, 28 x 3 = 84, repeat), as well as 27 and 29 (no valid value we could add to the product and stay under), leaving 26, which I tested. - 26 x 3 = 78 [1, 4, 5, 9]; 14 gives 92, failure, no other valid combinations for 2(?) values. - Of the x3 multiplier options with an initial value of 1(?), we've eliminated 13 and 15, and can quickly eliminate 12, 14 and 17 (products contain repeat digits), leaving 16, 18 and 19. - 19 can be eliminated because its product is odd, but it leaves only even digits to add, so the answer cannot be valid. - I didn't think of a quick way to eliminate these last logically, so tested them. - 16 x 3 = 48 [2, 5, 7, 9] Lead digit could only be 2. Second digit could only be 7 (answer uses 5) or 9 (answer uses 7), but sum will be in the 70s, so both are eliminated. - 18 x 3 = 54 [2, 6, 7, 9] Lead digit could only be 2. Second digit could also only be 2 (answer uses 6), so this set is eliminated. - Multiplier cannot be 3. Just x2 set to eliminate now. - If we start with a multiplier of x2, we would theoretically need to check the 1(?), 3(?) and 4(?) possibilities; 2(?) is out, and anything 5(?) and up is as well. - Any initial value ending in 6 is automatically eliminated, as it would leave us with a second 2. - Of the x2 multiplier options with an initial value of 4(?), we've already eliminated 41, 42, 44, 45 and 46. Any product over 85 will not leave digits that can keep us under 98, since our minimum addition would be 13 - which eliminates the rest as well. No need to test here. - Of the x2 multiplier options with an initial value of 3(?), we've already eliminated 31, 32, 33, 35 and 36, and can knock out 37 (product is 74, repeat) quickly as well. - We can also logically knock out 39 without testing, as the highest digit it leaves is 6, while its product is 78; we can't form a valid answer. - We can quickly eliminate 34 and 38 as well; their remaining digits means that the lead digit must be 1, to stay under 98, but this means the answer would be in the 80s, regardless of other combinations; we have no 8 for either, so they are eliminated. No need to test here. - Of the x2 multiplier options with an initial value of 1(?), we've already eliminated 11, 12, 15 and 16, and can eliminate 13 and 14 quickly without testing (products in the 20s, repeat). - I didn't think of any quick way to knock the remaining 3 out logically, so I tested them. - 17 x 2 = 34 [5, 6, 8, 9] Lead digit could only be 5, anything else goes over 98. Second digit could also only be 5. Eliminated. - 18 x 2 = 36 [4, 5, 7, 9] Second digit could only be 9 (answer uses 5); 49 gives 85, failure; 79 gives 105, failure. Eliminated. - 19 x 2 = 38 [4, 5, 6, 7] Lead digit could only be 4 or 5, Second digit could only be 6 (answer uses 4) or 7 (answer uses 5); 56 gives 94, failure, 47 gives 85, failure. Eliminated. - Multiplier cannot be 2. All other possible configurations have now been eliminated, thus 17 x 4 = 68 + 25 = 93 is the only possible solution. Q.E.D. ((For time-sense purposes, reasoning this out, with the small handful of quick scribbles that I tested, this process took about three minutes for me, which seems.... reasonable(?) without a 'clever' solution?))
@@santiagoruiz7490 Thank you ^.^ The irony is that it took so much longer to type up the explanation for the reasoning than thinking it through did... but it was a fun exercise!
Just from the video length, the solution process is clearly "try everything, with simple filtering of impossible cases". In that case, it's only a minor difference to convert that to "hey computer, try everything"
Yeah. After watching this for a minute I quickly wrote a python script to try all the permutations and got the unique solution before the video got to minute 5...
I paused and basically brute-forced the solution by trial and error, without noticing how long the video was. I quickly worked out that 1 can't be _b_ or _c,_ and 5 can't be _a, b_ or _c,_ so I basically assumed 1 is _a_ and 5 is _d_ (otherwise it gets hard for _h i_ to be a two-digit number), and just kept trying stuff, but I felt like I was missing some other way of actually proving where one of the digits was. I was getting stuck until I realised that _e_ + _g_ can be more than 10, resulting in a carry. Edit: Wow! I thought it might not be a unique solution when he found the solution I found, a third of the way into the video. Checking numbers in the 30s and 40s was insane! Any 3 _b_ × 2 is never going to work, because you'd always have to add 2 _g_ or 3 _g_ to it, to stay within two digits, and any 4 _b_ × 2 is either over 90, or doesn't leave any small digits to add to it. I can't believe he actually kept checking past 45, where you have to add to something that's already over 90, and somehow still get a two-digit number. Repeated digits are irrelevant at that point.
Here's some simple filtering: The single-digit multiplier cannot be 1 or 5 for the reasons presented in the video. If it's 6 or more, then the first digit of the other multiplier must be 1 (or the product is at least 120). But then the number added to the product is at least 23, and the final sum is at most 98, so the product cannot exceed the difference (98 - 23 = 75). The smallest product is 12 * 6 = 72, but that repeats the digit '2'. Any bigger number is also impossible: 6 * 13 = 78 > 75 and 7 * 12 = 84 > 75 and multiplication of positive numbers is monotonically increasing in both inputs. So the single-digit multiplier is either 2, 3 or 4. That's already some heavy pruning by a simple argument. For those possible multipliers, make lists of cases and remove any which results in 0s and/or duplicates. I did the list for 4 and parts of the list for 3, then got bored and watched the video. It was equally boring. I think my approach yields about as many cases as Presh's approach, just in columns instead of rows. Well, I have fewer cases of "the one past 100". So I guess work could be saved this way.
A small Python program : from itertools import permutations perm = permutations([1, 2, 3, 4, 5, 6, 7, 8, 9]) for p in list(perm): (a, b, c, d, e, f, g, h, i) = p if c*(10*a + b) == 10*d + e and 10*d + e + 10*f + g == 10*h + i : print(p)
To add to this, since the list given to permutations was sorted, the resulting perms will also be in sorted order. Once you see that a = 5, you can stop checking all future perms for the same reason Presh stopped at 50!
The code is not difficult to understand I guess the only smart part is the permutations and I think he did it himself, I wouldn’t have thought of it like that but Yh idk how to explain or idk what I am talking about
You can eliminate a ton of possibilities from the get go by noting that the multiplication has to be less than 76. Then, instead of cycling through constraints on the double digit number, look for constraints on the single digit. This gets you down to that number having to be 2,3 or 4. Then you can proceed checking those three iteratively
3 is easy to check, since there are only results of the multiplication that satisfy the constraints (repeated digits, less than 76). These are 51,42, and 72, but all Of these end up having repeated digits once you divide them by 3
@@herogpi1not bad thinking but once you see that the 8 is repeated, you’d have to do more subtraction to get something that would not result in a duplication or a zero.
Before adding the smallest combination of remaining numbers after the multiplication to check if it is greater than 99, a simple "is DE bigger than 86?" check would speed things up by a lot. You see, the highest number that HI can form is 98 (not 99), and on top of that, the smallest FG possible is 12, so it means that the largest DE possible is 86. And if you're checking arrangements with A = 1, the maximum DE is even lower because F can't be 1, so in this case, considering that the maximum HI is 98 and the minimum FG is 23, the maximum DE is 75. This also means that: if AB = 12, 2
This... plus remembering already established rules like "not multiplying by 5" or "don't repeat numbers" like 3x * 3. Even with bruteforcing this problem, half of the cases checked here could have been discarded in the first place. Don't know why it bothers me so much, but it does. ^^
A slight improvement to this brute force method is to realize that the addition of two 2-digit numbers will result in a 3rd 2-digit number. If the product resulting in the 1st 2-digit number is anything greater than 86, you don't have to check what the addition will be because the smallest 2nd 2-digit number you could add is 12 which results in 98 (the largest possible final result). This will quickly eliminate any 40's checks because 86 is the smallest possible product (43*2) that falls within the rules of the problem (any multiplier greater than 2 makes the product's 10's place at least 12).
Yeah, he kept going through cases that could’ve been eliminated, he also kept multiplying by numbers that already made smaller numbers too big for a couple of times before getting rid of them
86 can't be the result of the multiplication because you could only add 12 or 13 which would lead to 98 or 99 making the 8 or the 9 a repeated digit. I think just from the addition 85 would be the highest number that would work, but obviously 85 is not achieveable by multiplication without repeating a 5
@@Alvin853 you can push it even further, 84 doesn’t work because the only way that you have a five somewhere in the answer is if you add 11 or add 15 (because 5 can’t be in the multiplication), but 11 has a repeating one and adding 15 will give you 99 with a repeating nine, 83 doesn’t work because it’s prime, so 82 is actually the number you should be using as your lowest
@@ezracohen6020 if we're only looking at the addition step, then 83 would be possible with 83 + 12 = 95. But as you said, 83 can't be the result of a multiplication. If we keep going like that' we're doing the brute force in reverse though, 82 can only be achieved by doing 2*41 having a duplicate 2, 81 would need to be 27*3 which leaves no low digits for the addition (lowest digits left would be 4 and 5, 81 + 45 > 100), 80 isn't allowed, 79 would need a single digit number to be added to not have a 9 in the result so that's not allowed, 78 is the first one where you get multiple options
After seeing it was a 22-minute solution, I went to the comments to see if it would be worth while even trying. This comment helps me know not to. Thanks!
You can do a much shorter elimination, leading you to test out only a tiny handful of possibilities, if you focus on the multiplier first; grouping by multiplier sets (multiplier can only be 2, 3 or 4), then by lead digit after that, made it easy enough to eliminate out in my head in a few minutes with only a couple of quick scribbles to check.
I would give you a big and special thanks to you Presh because you helped me think more about how to solve problems like these, something I was unable to do some years ago in high school. I watch your videos since 2017 and I always find myself learning and improving. Thank you!
Aah! and there I was expecting some elegant mathematical solution. However, It taught me some logical thinking and that sometimes, the only way to solve a problem is to yomp though it. 🤔--->🤯--->🤕--->🤓👍
I looked for how I might solve this more analytically. The following steps were the best I came up with: 1. Prove that if de > 75 then either f = 1 or g = 1 2. Prove 1 < c < 5 3. List the possible combinations of a and c (three where a = 1, three where it isn't) 4. Prove d 5 7. Prove c = 4 8. Complete the solution A couple of steps (#4 and #7) require individual cases to be completely worked through but only a handful.
The full thing: 1. Prove that if de > 75 then either f = 1 or g = 1 Some combinations of d, e, f and g for de>75 are straightforwardly impossible. If de>88, de>80 and f>1, de>76 and fg>22, or de=76 and fg>23 then de+fg >= 100. If de=76 and fg=23 then hi=99, so h=i. If fg=22, then f=g. fg cannot be 20 as 0 is disallowed. As fg>=12, that means if de>75, fg must be between 12 and 19 or 21: f or g must be 1. 2. Prove 1 < c < 5 If c>6 or c=6 and ab>12, de>75 so f=1 or g=1, so the lowest value of ab must be 23, and ab * c >= 90 so de+fg >= 100. If c=6 and ab=12, de=72, so b = e. If c=5, e would either be 5 (clashing with c) or 0 (disallowed). If c=1, b=e. So 1 < c < 5. 3. List the possible combinations of a and c a must be less than 10/c else ab * c >= 100. a cannot be the same as c. If c=4 and a=2, then de>80 so f=1 or g=1, so ab >= 23, ab * c >= 90, and de+fg >= 100. Six options remain. If a=1, then c could be 2, 3 or 4. If a is not 1, c=2 and a=3, c=2 and a=4, or c=3 and a=2. 4. Prove d = 100. If d=8, then f=1 and h=9 so that de+fg 75 means f=1. There are three possibilities of a, b and c giving 75 < de < 80. If ab=38 and c=2 or ab=26 and c=3, then de=76 or de=78 respectively, h would need to be 9 as it must be greater than d and 8 is already taken, so 4 and 5 are left for g and i, which cannot work. If ab=39 and c=2, then de=78 and there is no option left for a value of h > d. For de = 100. 74 could only be created when c=2 and a=3, meaning ab=37 so b=d. 75 is impossible because e cannot be 5 unless b or c are also. Therefore, d = 100. So a = 1. 6. Prove b > 5 As a=1, if b * c < 10 (c=2 and b 10 and h = f+d+1, so f must be less than 5, the remaining digits are 2, 5, 7 and 9, so f=2 and h=7, but then g and i would have to be 5 and 9 which doesn't work. When c=3 and b=8, de=54, the remaining digits are 2, 6, 7 and 9, the only combination which fits e, g and i is g=2 and i=6 but then f and h would have to be 7 and 9 which doesn't work. When c=3 and b=9, de=57, the remaining digits are 2, 4, 6 and 8, and no combination fits g and i with e. So c=4. 8. Complete the solution If b=6 then e=4 which clashes with c. If b=8 or b=9 then d > 6. Therefore, b=7, so d=6, e=8. As h > d and 7 and 8 are taken, h=9. We can fit the remaining values, 2, 5 and 3 to f, g and i to find the solution: 17 * 4 = 68 + 25 = 93.
@Ranoake by this logic , hackers aren't bruteforcing passwords either because for example they know a minimum lenght on some website is let's say 7 charachters and it needs to have 3 numbers, so they skip all combinations under 7 charachters and with less than 3 numbers... they eliminate billions of combinations but its still bruteforcing it by checking each possibility...
Given letters to the squares from a to i from top to bottom: c1, because times 1 repeats the digits => contradicts different digits (cdd) c5, because times 5 terminates by 5 => cdd or 0 => not possible For the same reason, b 5 c9, because then 10a+b cdd c8, because then 10a+b too much c7, because then 10a+b too much for addition 13*7=91=>cdd 12*7=84, leaving 35 as the smallest to add => too much c6, because then 10a+b cdd 13=6=78, smallest to add = 24 => too much That leaves c = 4 or 3 or 2 c=4 => 10a+bcdd 13*4=52, leaves 67 to add => too much 14*4=>cdd 16*4=64=>cdd 17*4=68 68+23=91=>cdd 68+25=93 => !! SOLUTION !! 68+29=97=>cdd 68+32=100=> too much 18*4=72, leaves 35 to add => too much 19*4=76, leaves 35 to add => too much 21*4=84=>cdd 23=4=92=>cdd The rest of the reasoning is more of the same somewhat tedious case by case examination showing there's no other solution than the one we found.
d can't be 1( cause a must be 1 too), d can't be 2 ( c and a must be 2 and 1) and d can't be 9 ( because f >=1 , 90+10 >= 100). And we know, that de = ab*c, so de is not prime. Also we know that de and e5 I think easier to check all variants for de.
That's so much brute-force. First, just obvious check that -ab- < 44, because -de- can't be 9x because otherwise you can't add 2 digits number, brute force for the other 4 numbers to realized that the entire 4x row is invalid, narrow it down to -ab- < 40. Then, b cant be 1 or 5, as you said, and c cant be 1, 5 and 9. Brute force c by putting 8,7,6 on the 1x numbers and those weren't fit too, make the c possibility goes down to 2, 3 or 4. If you made a sheet outside before, there will be 2 -ab- "boxes", the one box that ended in 2, 3, 4, and the other box that ended in 6, 7, 8, 9. You can brute force the 234 box first, it's small enough and the c number will make the elimination faster, and then check the other box, which is just 12 numbers to do manually
I appreciate this channel, this channel is very good. I am from India and learned many ways to solve problems. I usually see video of this channel. This is a good ❤channel❤ for student.
I will say a slight shortcut to elimiting answer is that any possible that has d = 9 will not work as d+anything > 9 so there is hard lose cap of de < 90 (you can expand the logic a little more to say it has to be less then 85 because the small number you can therotically ever add is 12 and you can only go up to 97 because of repeat digits) I am sure there are other cool limitations to the varibles out there that i still have not thought of
yeah, h from get go should be greater than 6, because the smallest a and c are 1 * 2 => smallest d is 3 => smallest f cannot be 1 or 2, it should be at least 3 because for 1 and 2 have already been reserved => h is 6 or greater.
Did it pretty much the same way, except I took c as my base for it, since it only has 3 options. 1 is out for repetition, 5 for getting a 5/0, 6+ is out because the smallest option (13*6=78+24=102) is too big. Leaving 2, 3 and 4 as options. If c=4, A can't be 3+ obviously. So we need to check the remainder (12, 13, 16, 17, 18, 19, 23. Anything bigger overflows). 12 doubles, 16 doubles, 23 doubles, so 13/17/18/19 remain to check the way you did. If c=3, A can't be 4+, nor can it be 3. So we have to check 12,14,16,17,18,19,24,26,27,28,29. Without doubles, that's 16, 18, 19, 26, 27, 29. If c=2, then A can't be 2 or 5+, so it must be 1, 3 or 4. So we have to check 13, 14, 16, 17, 18, 19, 34, 36, 37, 38, 39, 43 (46+ makes 92+ so it'd overflow even with +10). 13, 14, 16, 36, 37 double, leaving 17, 18, 19, 34, 38, 39 and 43. So while it looks daunting, simple elimination only leaves us 4+6+7=17 options to bruteforce check. Which, once you get the hang of it, is pretty fast: options for f and h tend to be limited, if any exist; and g and i can sometimes be eliminated by an odds/evens problem outright. Also note that many of the options to check use the 7. 8 or 9, which eliminates key options for h. So it takes mainly diligence to check and find 17*4=68; +25=93
My solution did a lot more whittling before going brute force, on only 11 cases. Define letters as Presh did, so AB x C = DE, DE + FG = HI. As Presh noted, neither B nor C can be 1 nor 5, as they lead to re-use or a zero in E. Note that the minimum value for AB and FG are each 12. Likewise, the maximum for GI is 98. This means that the maximum DE is max(GI) - min(FG) = 86. As C >= 2, this implies AB =12, DE
I found a few more limitations...like "c can only be 2,3 or 4" and "a can only be 1 or 2". de cannot be prime, end in 5 or 0. This narrows it down to 24 options for de, with 30 corresponding possibilities for ab*c. It's easy to whittle those down, if you remove ab has to be 2 digits and a,b,c,d,e cannot be repeated. I got 11 such combinations and then brute checked those.
I knew it would be A LOT FASTER to solve this problem by the kind of method used by Presh but I couldn't resist trying to solve it with an Excel spreadsheet. It took "a few" hours to create, but it worked.
Notating top left as A5 and bottom right being B1. a few obvious logic points, A5 has to be 1-4, (likely 1 or 2). B4 has to be 2-4, A1 has to be 6-9 (likely 8 or 9). 5 can't be in the multiplication digits (gives 0 or 5 as a product) etc. Other points- B5 is a nice place to put a big number, as is B3. I don't think I tried more than 10 permutations before I got the answer.
A way to reduce the verifications if the sum is bigger than 100: Suppose the fifth number is 98 and the fourth number is 12, then the maximum value for the third is 98-12 =86. Then, we search only three values (a,b,c) such (10a+b)*c
Damn I didn’t see this comment. That’s exactly how I did it. It’s much easier to prove c can’t be high numbers and limit it to 234, and work from there.
This was a horrible problem requiring brute force, I appreciate your patience with "big" numbers because a "xx+12" surprise solution may have existed. I was hoping for a math equation system, but my LIKE goes to the editing quality.
If you are a programmer, this is one case where a little light recursion can save you a whole lot of grief. Nine nested for loops is not a pretty construct. Compare it with this pseudocode: def test_sum (arguments) if (we have exactly 9 digits) then if (the sum is satisfied) then print the answer else just exit fi else for (x = each digit that has not been mentioned) test_sum arguments, x next fi fed test_sum 1
In the multiplication section you can eliminate any number over 90, since the addition number must be 2 digits. This also lets us stop checking at 45x2.
@@richbuckingham we can probably make it smaller still by arguing that anything over 33 can only be multiplied by 2, thus the addition must be 13 or greater and as you said 98 is the max final value so 85 is the multiplied total, but the multiplication can’t end in a 5, so 84. So we can stop checking at 42x2, which is obviously wrong, so 39x2 would be the last real check.
I tried the questions before myself, I didn't know the technique he used in this video yet i got the first solution (17 x 4). I was stunned for a moment
You can also do this backwards: picking the highest result number (98), checking for possible subtractions, and eliminating repetitions and indivisible numbers.
Checking most combinations like in the video is a valid way to do this There are clever eliminations that can be used 1. Noticing we must add another 2digit number, so the product 20 afterward) check: 4 21,23 don't work. So 12~19*4=48~76, eliminate wrong combinations, left 52 68 72 76 (72+3X>99, 76+2X>85) brute force through 52,68 to find answer 17*4+25=93 check:3 if we use 20s to multiply 3, the equation should be 2A * 3 + 1B = CD (obviously cannot add 4X) looking at multiplication, only 26*3=78 and 29*3=87 works, checking the addition, no answers here 1A * 3 + BC = DE multiplication: only 16*3=48, 18*3=54, 19*3=57 works, brute force to find no answers (Brute force, start from the smallest) check: 2 30s, 3A * 2 + 1B = CD 34*2=68, 38*2=76, 39*2=78 (68+1X=8Y, fail; 76+14or15 fail; 78+1X
I used a similar start to eliminate 1 and 5 from being B or C. However, I then showed that C could never be greater than 5 because if C is greater than 5, A would have to be 1 (otherwise the multiplication would have a 3-digit result). 12x6=72 and 72+34>100. Similarly, 13x6=78 and 78+24>100, and anything else just results in an even larger 3-digit number after the addition so C could only be 2, 3, or 4. After that, it was easy to prove through brute force that the only solution was 17x4=68 and 68+25=93 similar to how you did.
Another cool trick: Any product that is 88 or above can be eliminated because you have to add a 2 digit number to it that would make the result 3 digits and the smallest possible 2 digit number without repeating is 12
I think you can also eliminate numbers that give you higher than 84 after multiplication. You need to add a two digits number, and the lowest one possible to add is 12. The number after addition can not go above 98, so 98-12=86. But in this case there is a repeated 8, so the number in question goes 1 lower, and we can't have a number with a 5 on the end, so our final number is 84. This could eliminate quite a few possibilities without a need to check them. For example you don't need to multiply by 4 any number in 20-s
I love solving for a solution based on the thumbnail before watching these videos for alternate perspectives. Surprised to see only one for this problem, and error checking isn’t that great
1,7,4,6,8,2,5,9,3 f(from a to i). This has nothing to do with brute force. It's a successive process of elimination. After investigating combinations with a high likely hood of digits being generate multiple times, and zeros being produced, they were dropped. The main contestants for the initial multiplication (contained in the first 2 fields) are therefore either 17, 18, or 19. The multiplier in the 3rd field can only be 2,3, or 4. 4 bombed out immediately, one case produced a double digit and one went past the boundary of 98. That left merely 7 combinations with the first 5 digits being in order, plus 4 defined digits to play around with. Easy as pie. You combine 2 of those leftover digits and add it to what's already known and see if the result matches any combination of the remaining two digits. You only have to do that 7 times! 6 times I failed and on the very last option left being (17x4=68), voila! 68+25=93.
Commenting before seeing the solution: I've brute forced in my head and got 17x4=68+25=93, I wondered if there's any more solutions. Update after watching: Let's FREAKING go. I'm not joking, I literally worked out the one and only solution, without any papers and pens. That's incredible!!! 😊 Bonus: There's a few more equalities of the same kind that I know on top of my head 54 x 3 = 27 x 6 = 18 x 9 29 x 6 = 58 x 3 = 174
I did much the same. As long as you have the right initial goals in mind it's not too hard to get it quickly. In addition to the stuff with 1 and 5, you know the last 2-digit number is going to be the highest, so you want to try to avoid using any big numbers (6, 7, 8, 9) in your initial multipliers so you can add up to 80- or 90-something at the end, and you want to have a low 10's digit (2, 3) in the second to last number so that the multiplication result doesn't have to be too low. 1 seems like the ideal 10's digit for the multiplication, since with 20 the products start to get big enough that there's no space left before 100, and you don't want to use two odd numbers for the other two because you know you'll probably need the 9 for the last number. Same with using both even numbers, since you're already guaranteed to use at least one more even number for your product if either of your multipliers is even. All that together, and basically something-teen (odd) times 4 is probably the best place to start. And there are only so many options (basically just 13 and 17). As soon as I saw 68 for the product, before I even considered any actual further calculation, I knew I probably had it, because that's all the large numbers gone but 9 (which I expected to use as the final 10's digit anyway), and 2 still available so I can get to that 90-something sum with ease.
The part where you say any number over 98, I would say cut any number in the multiplication over 85 as the smallest 2 digit number is 12 that also leaves you with a 2 digit number at the end that does not repeat any with 8 is 97. Would help with the multiplication part.
Neat puzzle. After some reasoning and doodling on a scratch pad for 2-3 minutes, I had concluded that c must equal 2, 3, or 4 and that a must equal 1, 2, 3, or 4. From there I did go to a more "brute force / eliminate possible answer" approach and came up with the solution. I had worked in the opposite order as Presh, tackling a = 4 first, then a = 3, et cetera since those have much more restrictions on what c could be. In my opinion, this 22 minute video could have been more like 9-10 minutes, if a bit more time had been spent on the "front end" of eliminating possible digit combinations. As others have said, too much of a "try all combinations" approach ultimately isn't a very satisfying strategy.
"In a remote and mysterious land, there exists a temple hidden deep within a dense jungle. The temple's entrance is guarded by a series of ancient statues placed at strategic locations. One particular statue, known as the Guardian, casts a shadow that holds the key to unlocking the temple's secrets. The Guardian's shadow is cast on a sacred stone platform, creating a triangle of shadow. The temple guardians believe that the shadow triangle is similar to a smaller triangle inscribed on the stone platform. The smaller triangle represents a map leading to the temple's inner chambers. The lengths of the sides of the shadow triangle are given by the quadratic equation s^2 - 6s + 9 = 0, where s represents the length of one side of the shadow triangle in meters. Similarly, the lengths of the sides of the smaller triangle are given by the quadratic equation t^2 - 4t + 4 = 0, where t represents the length of one side of the smaller triangle in meters. To decipher the temple's secrets, the guardians must determine the ratio of the perimeter of the shadow triangle to the perimeter of the smaller triangle. Find the ratio A) 3:2 B) 4:6 C)1:8 D) 2:5 Find the correct answer
the smallest number you can make for the addition is 12, so no value larger than 87 (because 88 is double) from the multiplication has to be checked. saves some minor time of this bruteforcing
Wow, this has to be one of the most brute force Presh videos I've seen. I was also able to find the unique answer but I had a different approach. I focused on 'c' and I was able to prove easily enough that it had to be 2, 3 or 4. So I would then consider the possible values of a and b for each c, and work through each one. I tried c=4 first and found the solution (and the only one that worked for c=4), but I then proved there were no solutions for c= 3 and c=2. I feel my approach might have been a bit faster but that's maybe up for debate. Still requires a lot of checking.
Normally, I just proceed with the video because I can't solve it. This time I did and I'm glad I did because the video just checked everything instead of skimming the possibilities through logical eliminations
My intuition was that "c", being a multiplier on a two-digit number that must result in a two-digit product, would be the most restricted. So I tried each digit 1-9 as "c", and found that most of the digits led to immediate dead ends. c=1 obviously forced duplicate digits. c=5 forced either a duplicate e=5 or e=0. c≥6 forced three-digit sums at minimum. So "c" must be 2, 3, or 4. Then, for each c=2, c=3, c=4, I checked all possible "b", and computed "e" modulo 10. This led to only 16 b-c-e triplets with no 0s and no repeated digits: 3x2=6, 4x2=8, 7x2=4, 8x2=6, 9x2=8, 2x3=6, 4x3=2, 6x3=8, 7x3=1, 8x3=4, 9x3=7, 2x4=8, 3x4=2, 7x4=8, 8x4=2, 9x4=6 (all modulo 10). 16 candidates seemed a reasonable amount to brute force, so brute force I did. Each triplet has few, if any, valid values for "a", and "d" can be computed. So it's not much effort to check each triplet. Only the 7x4=8 triplet leads to a solution.
my thought was to create the largest Sum answer you could and then work backwards, since the final answer is the Sum, start with 98 and figure out how to make a sum of 98 with the remaining numbers and work your way down from there.
You could have stopped sooner by keeping in mind that the final step requires a 2-digit non-repeating number which still adds to 98 or less since 99 is repeating. So the final step can't be 1-9/10/11 so the max middle result is 86, assuming 12 for the last step. Although 12-16 would be eliminated too due to repeating 8 for 98, they might work at lower totals. So as soon as 43x2, using all the lowest factors in the first 3 blocks, came out to 86 and the 2 was a repeat, anything past that would have forced an end result above 98. Work it from both ends.
17x4=68 => 68+25=93 :) I went by testing with the lowest possible values for a and b. I also eliminated impossible solutions like c=5 and b=5 for example.
I did the same general process but I eliminated more options for C before going through the teens. 9-6 got eliminated, because they were out of bounds. The best case scenario would be 13*c+24
Here is how I found the right answer (I think much more quickly) I considered the single digit multiplier first. That combined with the two-digit result imposes a major limitation. Second thing to check is the smallest remaining digit to place in the tens digit of the added number. 12 is the smallest multiplier, then 34 is the smallest amount you can add. 9x12 too large 8x12 too large 7x12 too large 6x12 = repeat 2 6x13 = 78 +24 = too large 5 is impossible because of 5/0 ending 4x12 = repeat 4 4x13 = all the small digits used (52, the 10s digit of the added number needs to repeat) 4x16 = repeat 4 4x17 = solution
Brute forcing the combinations would be trivial on a computer, but the algorithm might look cooler if it used a few tricks that humans might apply on paper. Upper bounds on hi (98) and lower bounds on fg (12) to limit the range of de to a maximum of 84, plus some divisibility rules and using sets in combination to manage the remaining options so the whole process looks concise. It was fun on paper.
I was surprised you kept going so long. The first thing I did was work backwards and determined the largest possible number at the end is 98. So my equation looks like this: ab*c+fg=98 Next, the smallest number that fg could be is 12. ab*c+12=98 -> ab*c=86 The smallest number for c is 2 (yes it is against the rules, but ok for making a bound) ab*2=86 -> ab = 43 So 43 is the absolute maximum that I could quickly prove that could work. No need to go all the way to 50.
Instead of checking for 99, you could've eliminated the multiplied values that are greater than 86 itself as a two-digit number of at least 12 is to be added to it to arrive at the final value of less than 99.
Can take it a step further to 85 since 86+12=98, where 8 is repeated and can’t be a solution. Edit: as another commenter pointed out, it can’t end in 5 so 84 is the highest possible.
@@zxbot4718 It's a minimum of 12 and not exactly 12. But nos 10 and 11 are not of that case, since the former has zero in it and the latter has repeated nos.
Nice puzzle. But you could have stopped with the maths way earlier. The biggest number in the end could only be 98. And you have do add a 2 digit number to it, which is at least 12... therefore the the difference is at ma 86... This 86 is the maximum for the product (and not 99). - But I understand your solution for this video to keep the numbers simple.
You can ignore all cases resulting over 86. The smallest number you can add is 12. 87 + 12 is 99 which is already too big. So based on that there are lot more cases you didn't have to check
This is brute force, but it’s also a very bad brute force. Although the calculations are minimal, a lot of time is wasted checking digits against 6/7/8/9 for c. As you said, 1 can’t be c, nor 5. But given the sum adds to a 2 digit number, 9 is out since it’s too big, 8 same reason, 7 has a chance with 12 but that’s 84 and you can’t add a 1z number. 6 loses 12 with 72, 13 is 78 but the smallest is 24, and 14 has the same problem as 7x12. Now you’ve eliminated 1,5,6,7,8,9 for c. 4 is fairly simple as the upper limit on it is 4x19, and you can find 4x17 fairly easily. If you’re told this has 1 solution, this is the fastest way to deductively solve. If you need to prove it, you can eliminate the 4s this way, the 3s have an upper limit at 29, and the 2s at 44. 46-49 are needless checks, and any multiplying of 2 or 3 after their limits is unnecessary.
The approach could have been simplified significantly by saying that the multiplication result cannot be higher than 85, because 85 + the smallest allowable two digit number would be above 97 which is the highest allowable result when dealing with the 80s.
I found a sort of analytical approach but with lots of assumptions: Lets rewrite the question to ABxC=DE + FG=HI First, looking at the one's digit, there are 6 possible cases: 1. even×even=even +even=even 2. even×even=even +odd=odd 3. odd×even=even +even=even 4. odd×even=even +odd=odd 5. evenxodd=even +even=even 6. evenxodd=even +odd=odd Second, we can note that there are more odds(1,3,5,7,9) than even(2,4,6,8). Therefore, we can compare each cases and assume we want the one with the most odds/least even. (Cases 4 and 6) Another assumption is that we'll start with a small number and end with a large number. Thus, we will assume the A=1 and H=9. After all these assumptions you can reach the answer by yourself. The lines after this is for those who are stuck or want to continue the solution: We can note all the locations of the odd numbers (A,B or C,G,H,I) and determine the location for even (B or C,D,E,F). Next, since D and F are both even. The only way they can be 9 is if D+F=8 and then E+G>10 So, D and F are either 2 or 6 E is even, so it is either 4 or 8 Cases for E+G>10: If E is 4: G is 7 but 4+7=11 and 1 is already used This means E is 8: 8+3=11 (can't be because 1 is already A) 8+5=13 which is possible 8+7=15 which is possible In both possibles cases 5 is being used. Looking at the one's digit: BxC=E(8) Options remaining are 3,4,7 Only possible cases are 4×7 or 7×4 Therefore, I=3 and G=5 Since B and C are either 4 or 7: D cannot be 2 and must be 6, F=2. Thus ABxC=DE is 1(4 or7)×(7 or 4)=68 Which makes B=7 and C=4 Again, lots of assumptions but a more funner way than brute force.
instead of just eliminating results over 99, you can immediately eliminate any result where the multiplication is greater than 89. There is no 2 digit number possible to add to 90 to get back a 2 digit number.
Through setting up a main logic grid, and then a smaller one once your C possibilities are dramatically reduced, you can narrow it down to about 12 possible answers for the "DE" product. Then just check those with the remaining digits. You checked so many unnecessary cases.
My thought process: maximum final result is 98, since its 2 digits and cannot repeat the 9. d must be less than 9 so any multiplication over 90 can be discarded c cannot be 1 as that means a=d and b=e. and b cannot be 1 as that would make c=e this limits a to
I only checked for C term and started with the largest possibility and slowly went down I reached from 9,8,7,6,5 and then when I checked for 4, it worked out. I started checking with 12, 13, 17 and found the answer
I found this solution too but was hoping there was a more elegant solution than check all viable options. Guess not. I did eliminate intermediate values at 86 though instead of 99 since the minimum number you can add is 12 and the max valid result is 98.
use a little logic to try to figure out the top, and it was a bit bruteforce at the end but got there. I mostly was like "top has to be small, bottom probably has a 9 as a target, and b&c cant be 1&5"
This problem might have 2 interpretations. 1. This could be a 2 step multiplication algorithm with the final answer on the bottom. 2. It could be interpreted like in this video
I kinda worked backwards a little. I knew, based on the criteria, that the final solution could be no larger than 98. In order to get a final solution that was no larger than 98, the multiplication solution had to be no larger than 76. (I knew this because 98, the largest possible double digit combo, minus 12, the smallest possible double digit combo, is 86. But 86 would repeat the 8, as would anything down to 80. 79 would repeat the 9, 78 would repeat the 8, and 77 would repeat the 7.) This meant that my multiplier (the one single digit number) can't be anything higher than a 6. Again, the smallest possible double digit combo is 12, and 12 times 6 is 72; any larger multiplier (be it 13 times 6 or 12 times 7) would result in a multiplication solution higher than 76. But 12 times 6 isn't right, because 72 would repeat the 2. I also knew, for the same reasons the video stated, that the multiplier couldn't be either 1 or 5. That eliminated my multiplier to either 2, 3, or 4. So I started with the multiplier 4, as that would have the fewest multiplications I would have to check. Keeping in mind that my multiplication solution can't be any higher than 76, I divided 76 by 4 and found that it's a clean 19. But that wouldn't work because the smallest double digit addition possibility was 23, and 23 plus 76 was 99, which would not only repeat the 9 but also be higher than the final solution of 98. So I moved on to 4 times 18, which is 72. But that wasn't right because the smallest double digit addition possibility was 35, and 35 plus 72 is 107, which is higher than 98. Then I moved on to 4 times 17, which is 68. The smallest double digit addition possibility was 23, and 23 plus 68 was 91, which repeated the 1. The next double digit addition possibility was 25, and 25 plus 68 was 93. And this was the correct answer. I didn't time myself on this process, but it didn't take as long as this video did.
You keep eliminating anything over 99 but you could use anything over 88 - the smallest 2 digit number is 12. In practice, for the cases where you are using 1, the smallest 2 digit possible is 23 and you could eliminate anything over 77.
I don't want to be petty, but I see a better way for constraining the numbers are possible. As 0 is not allowed if the number (of the two added) is greater than 90, you will have to add a number less than 10. So that is not allowed. This (to me) is a better filter for the resultant addition.
I assumed the first digit would be a 1 or a 2 and the third would also be low. Guessed it very quickly making similar assumptions, like at least one of the 8 or 9 would be in a singles column in the addition section and the solution would likely start with 8 or 9. Wanting to save low digits for later, guessed 17 in the first spot. After stupidly trying 17 x 3 and seeing I already had a repeat, tried 17 x 4. With a last digit of 8, saw the 5 in the digit column of the next number would give the 3 and the rest fell into place. Bottom line is that I got lucky and got it with just that first 17x3 error.^.
no need to check after 43.. because plus number is more than 12 if you multiply 46x2 = 92.. then you need to only 7 to get 99... so it is not possible as you need to add 07 and zero is not an option..
Well that was disappointing. I was just trying to solve it from the thumbnail and used logic to eliminate several possibilities, but was still left with a lot of options. I figured there had to be some other logical deduction I was missing, but it turns out this was just a 'run through the remaining options until you find something that works' problem. More of a tedious waste of time than a fun puzzle. It's cool that it only had one solution, but extremely uninteresting in how one must find that solution.
This is much shorter to brute force:: first thing: *c cant be 1* ,so c is 2 or higher, thus *a cant be 9,8,7,6,5* which will result in 3digit number. Also *b cant be 5* as if it was-then e will be 0 or another 5 which both cant happen, *c cant be 5 as well* if it was-then e will again be 0 or another 5 which both cant happen. *e cant be 5* if it was,then either b or c should be 5. Now with this thing: go through every possibility of e from 1 to 9(5 cant come) the above conditions such as *a cant be 9,8,7,6,5* eliminates many possibility. We end up with e=8, i.e. 17*4=68, now with remaining 2,5,3,9 we know f is 2 or 3 else it will give 3 digit sum,and we discover that it is 17*4=68,68+25=93
Python solves it in a fraction of a second :) from itertools import permutations for digits in permutations(range(1,10)): if (digits[0]*10 + digits[1]) * digits[2] == digits[3]*10 + digits[4] \ and digits[3]*10 + digits[4] + digits[5]*10 + digits[6] == digits[7]*10 + digits[8]: result = digits break print(result)
Your references to "99" should be references to 98, because 99 already contains a repeated "9". The initial multiplication cannot add up to more than 86, because the minimum later addition is 12. That means no initial number above 43 works. That also rules out 43 as 2 would be repeated, so much of the tail end analysis could have been omitted.
Can we get Wolfram to solve it? tried to tell it: set {1,2,3,4,5,6,7,8,9} = {a,b,c,d,e,f,g,h,i}; 10ac + bc = 10d+e; 10d + e + 10f + g = 10h + i but it just asigned the a-i in order :-(
Yes, tried by hand first, got a few "that number can't go there", but run out of ides, and didn't want to brute-force it by hand, so did it in php. then watched your video, and noticed you to kind of brute-forced it in the end. Then I started to wonder how to solve it in wolfram.
I played around with it too. It's hard to say what syntax wolfram alpha will take to understand the question. It didn't seem to like the set equality statement in general.
Well, that was a very brute-force approach to this problem. I feel like we should let a computer do this kind of repetitive work if we're going to go through this much trouble. I wonder what elegant methods may get us here? Some were mentioned early knocking out possibilities.
Yeah... I was expecting some kind of cool solution myself, but this was pretty much brute force. Is like showing us a maze, and asking "How could we solve this maze?" and then it starts drawing all the paths to see if it reaches the end of the maze at some point. XD
Yeah, I came here hoping for an elegant solution. Instead, he persisted long beyond the point I was sure there were no other answers. SMH
could of course be made waaay faster but this suffices for me for a quick and dirty solution...
I think the computation could have been reduced pretty quickly by noting how big the product could be, considering you have to add a 2 digit number to that and still get a 2 digit number.
@@DawnDavidson hahahaha. The other guys were s1lly stup1d, but you were the one that made me laugh and feel the need to make this comment.
So, my mental reasoning for this one went:
- Final result must be under 100
- We don't have a zero
- Number after multiplication must leave us digits that can add to 98 or below
You can eliminate the vast majority of options from this, so what next...
- Multiplier cannot be 1; it would cause repeat digits.
- Multiplier cannot be five; it would cause repeat digits.
- Multiplier cannot be above five, as this doesn't leave any way of leaving you a second two-digit number that will keep you under 100.
- ((Multiplier must be 2, 3 or 4))
- Second digit of initial value cannot be 5; it would cause repeats, or a zero.
- Second digit of initial value cannot be 1; any result would require another 1 and be invalid.
- First digit in product cannot be 1; this is impossible mathematically (we are multiplying at least a two digit number by at least 2).
- First digit in the final answer also cannot be 1, for the same reason.
- Second digit in product could only by 1 if the multiplier is 3 AND second digit in initial value is 7; no other combination allows this.
(at this point I'd mentally eliminated a lot, but changed tack slightly without an immediate path of eliminating to a definite position for the 1)
- If the multiplier must be 2, 3 or 4, we can begin checking some combinations, because we'll quickly be able to eliminate whole branches of inquiry without needing to check.
- If we start with a multiplier of 4, we can only possibly have to check initial values in the 1(?) and 2(?) range - anything higher prevents us staying 98 or under.
- Of the x4 multiplier options, we can eliminate any combinations where the second digit is 4.
- Of the x4 multiplier options with an initial value of 2(?), we eliminate 21, 22, 24 and anything over 24 (product over 98) automatically. 23 can also be eliminated (4 x 3 = 12, 2 is used)
- Of the x4 multiplier options with an initial value of 1(?), 12, 14, 15 and 16, are eliminated automatically for repeat digits, leaving just 13, 17, 18 and 19.
- 13, 18 and 19 all leave us with remaining digits for which even the smallest possible addition will take us over 98. (52 with [6, 7, 8, 9] left, 72 with [3, 5, 6, 9] left and 76 with [2, 3, 5, 8] left)
- I couldn't think of an immediate way to eliminate 17, so I checked it manually.
- 17 x 4 = 68 [2, 3, 5, 9] Lead digit could only be 2, as anything else goes over 98. Second digit could only be 5, as anything else leaves an invalid unit digit.
- 17 x 4 = 68 + 25 = 93 - Success!
1 Valid Solution, then, seems to be: 17 x 4 = 68 + 25 = 93
==
I've been told there's only one solution, and this solution is valid, but I've not *proven* that there is only one solution; my tester might be mistaken, so I continued.
We've eliminated all other x4 multipliers, so the only place where another solution could exist would be in the x2 and x3 sets.
Still working down, I started thinking about what else I could eliminate from the x3 set.
- Of the x3 multiplier options, we can eliminate all initial values over 2(?) as well, just like for x4 multipliers; we can't use 3(?) for repeat digits.
- If we start with a multiplier of x3, we can only possibly have to check initial values in the 1(?) and 2(?) range; we can't use 3(?), and 4(?) puts us over.
- Of the x3 multiplier options, we eliminate any combination where the second digit is 3.
- Of the x3 multiplier options with an initial value of 2(?), we've already eliminated 21, 22, 23 and 25, and can quickly eliminate 24 and 28 (3 x 4 = 12, repeat, 28 x 3 = 84, repeat), as well as 27 and 29 (no valid value we could add to the product and stay under), leaving 26, which I tested.
- 26 x 3 = 78 [1, 4, 5, 9]; 14 gives 92, failure, no other valid combinations for 2(?) values.
- Of the x3 multiplier options with an initial value of 1(?), we've eliminated 13 and 15, and can quickly eliminate 12, 14 and 17 (products contain repeat digits), leaving 16, 18 and 19.
- 19 can be eliminated because its product is odd, but it leaves only even digits to add, so the answer cannot be valid.
- I didn't think of a quick way to eliminate these last logically, so tested them.
- 16 x 3 = 48 [2, 5, 7, 9] Lead digit could only be 2. Second digit could only be 7 (answer uses 5) or 9 (answer uses 7), but sum will be in the 70s, so both are eliminated.
- 18 x 3 = 54 [2, 6, 7, 9] Lead digit could only be 2. Second digit could also only be 2 (answer uses 6), so this set is eliminated.
- Multiplier cannot be 3.
Just x2 set to eliminate now.
- If we start with a multiplier of x2, we would theoretically need to check the 1(?), 3(?) and 4(?) possibilities; 2(?) is out, and anything 5(?) and up is as well.
- Any initial value ending in 6 is automatically eliminated, as it would leave us with a second 2.
- Of the x2 multiplier options with an initial value of 4(?), we've already eliminated 41, 42, 44, 45 and 46. Any product over 85 will not leave digits that can keep us under 98, since our minimum addition would be 13 - which eliminates the rest as well. No need to test here.
- Of the x2 multiplier options with an initial value of 3(?), we've already eliminated 31, 32, 33, 35 and 36, and can knock out 37 (product is 74, repeat) quickly as well.
- We can also logically knock out 39 without testing, as the highest digit it leaves is 6, while its product is 78; we can't form a valid answer.
- We can quickly eliminate 34 and 38 as well; their remaining digits means that the lead digit must be 1, to stay under 98, but this means the answer would be in the 80s, regardless of other combinations; we have no 8 for either, so they are eliminated. No need to test here.
- Of the x2 multiplier options with an initial value of 1(?), we've already eliminated 11, 12, 15 and 16, and can eliminate 13 and 14 quickly without testing (products in the 20s, repeat).
- I didn't think of any quick way to knock the remaining 3 out logically, so I tested them.
- 17 x 2 = 34 [5, 6, 8, 9] Lead digit could only be 5, anything else goes over 98. Second digit could also only be 5. Eliminated.
- 18 x 2 = 36 [4, 5, 7, 9] Second digit could only be 9 (answer uses 5); 49 gives 85, failure; 79 gives 105, failure. Eliminated.
- 19 x 2 = 38 [4, 5, 6, 7] Lead digit could only be 4 or 5, Second digit could only be 6 (answer uses 4) or 7 (answer uses 5); 56 gives 94, failure, 47 gives 85, failure. Eliminated.
- Multiplier cannot be 2.
All other possible configurations have now been eliminated, thus 17 x 4 = 68 + 25 = 93 is the only possible solution.
Q.E.D.
((For time-sense purposes, reasoning this out, with the small handful of quick scribbles that I tested, this process took about three minutes for me, which seems.... reasonable(?) without a 'clever' solution?))
This is the longest comment I've ever seen on TH-cam. Loved your reasoning and process of elimination!
@@santiagoruiz7490 Thank you ^.^ The irony is that it took so much longer to type up the explanation for the reasoning than thinking it through did... but it was a fun exercise!
replying to this message so I can read it later
I feel your answer needs to be pinned
Clever
Just from the video length, the solution process is clearly "try everything, with simple filtering of impossible cases". In that case, it's only a minor difference to convert that to "hey computer, try everything"
Yeah. After watching this for a minute I quickly wrote a python script to try all the permutations and got the unique solution before the video got to minute 5...
I paused and basically brute-forced the solution by trial and error, without noticing how long the video was. I quickly worked out that 1 can't be _b_ or _c,_ and 5 can't be _a, b_ or _c,_ so I basically assumed 1 is _a_ and 5 is _d_ (otherwise it gets hard for _h i_ to be a two-digit number), and just kept trying stuff, but I felt like I was missing some other way of actually proving where one of the digits was. I was getting stuck until I realised that _e_ + _g_ can be more than 10, resulting in a carry.
Edit: Wow! I thought it might not be a unique solution when he found the solution I found, a third of the way into the video. Checking numbers in the 30s and 40s was insane! Any 3 _b_ × 2 is never going to work, because you'd always have to add 2 _g_ or 3 _g_ to it, to stay within two digits, and any 4 _b_ × 2 is either over 90, or doesn't leave any small digits to add to it. I can't believe he actually kept checking past 45, where you have to add to something that's already over 90, and somehow still get a two-digit number. Repeated digits are irrelevant at that point.
@@Jivvi "you'd always have to add 2_g_ or 3_g_ to it to stay within two digits" - don't know how you ruled out adding 1_g_
@@richardtaylor4258I quickly went through my head and found a solution in less than that time.
Here's some simple filtering:
The single-digit multiplier cannot be 1 or 5 for the reasons presented in the video.
If it's 6 or more, then the first digit of the other multiplier must be 1 (or the product is at least 120).
But then the number added to the product is at least 23, and the final sum is at most 98, so the product cannot exceed the difference (98 - 23 = 75).
The smallest product is 12 * 6 = 72, but that repeats the digit '2'. Any bigger number is also impossible: 6 * 13 = 78 > 75 and 7 * 12 = 84 > 75 and multiplication of positive numbers is monotonically increasing in both inputs.
So the single-digit multiplier is either 2, 3 or 4. That's already some heavy pruning by a simple argument.
For those possible multipliers, make lists of cases and remove any which results in 0s and/or duplicates. I did the list for 4 and parts of the list for 3, then got bored and watched the video. It was equally boring.
I think my approach yields about as many cases as Presh's approach, just in columns instead of rows. Well, I have fewer cases of "the one past 100". So I guess work could be saved this way.
A small Python program :
from itertools import permutations
perm = permutations([1, 2, 3, 4, 5, 6, 7, 8, 9])
for p in list(perm):
(a, b, c, d, e, f, g, h, i) = p
if c*(10*a + b) == 10*d + e and 10*d + e + 10*f + g == 10*h + i :
print(p)
Thank you. This is a compact and sort of elegant way of solving the problem, and a good and inspiring demonstration of Python.
To add to this, since the list given to permutations was sorted, the resulting perms will also be in sorted order. Once you see that a = 5, you can stop checking all future perms for the same reason Presh stopped at 50!
_mathematicians:_ trying their ass off
meanwhile..
_programmers:_ hold my beer... there is an algoritm for this
Did you do that yourself? It’s published a lot on the web. If you did, it’s REALLY impressive, I have sooo much trouble thinking in code like that.
The code is not difficult to understand I guess the only smart part is the permutations and I think he did it himself, I wouldn’t have thought of it like that but Yh idk how to explain or idk what I am talking about
You can eliminate a ton of possibilities from the get go by noting that the multiplication has to be less than 76. Then, instead of cycling through constraints on the double digit number, look for constraints on the single digit. This gets you down to that number having to be 2,3 or 4. Then you can proceed checking those three iteratively
3 is easy to check, since there are only results of the multiplication that satisfy the constraints (repeated digits, less than 76). These are 51,42, and 72, but all
Of these end up having repeated digits once you divide them by 3
Why 76? Wouldn't be 98-12=86?
@@herogpi1 it’s the impact on the other digits. If you are adding 12, then the smallest the multiplication can be is (35)*4 = 140 (as an example)
@@herogpi18 would be repeated
@@herogpi1not bad thinking but once you see that the 8 is repeated, you’d have to do more subtraction to get something that would not result in a duplication or a zero.
Really cool that for a puzzle with 362,800 possible arrangements, there's only one possible solution. Fun puzzle.
*362880
You seen cluey. Can it be done with algebraic equations?
@@SpatialGuy77 I think so
You must have a better way. Trial should not be the only option
9!
Before adding the smallest combination of remaining numbers after the multiplication to check if it is greater than 99, a simple "is DE bigger than 86?" check would speed things up by a lot.
You see, the highest number that HI can form is 98 (not 99), and on top of that, the smallest FG possible is 12, so it means that the largest DE possible is 86. And if you're checking arrangements with A = 1, the maximum DE is even lower because F can't be 1, so in this case, considering that the maximum HI is 98 and the minimum FG is 23, the maximum DE is 75.
This also means that:
if AB = 12, 2
This... plus remembering already established rules like "not multiplying by 5" or "don't repeat numbers" like 3x * 3.
Even with bruteforcing this problem, half of the cases checked here could have been discarded in the first place.
Don't know why it bothers me so much, but it does. ^^
A slight improvement to this brute force method is to realize that the addition of two 2-digit numbers will result in a 3rd 2-digit number. If the product resulting in the 1st 2-digit number is anything greater than 86, you don't have to check what the addition will be because the smallest 2nd 2-digit number you could add is 12 which results in 98 (the largest possible final result). This will quickly eliminate any 40's checks because 86 is the smallest possible product (43*2) that falls within the rules of the problem (any multiplier greater than 2 makes the product's 10's place at least 12).
Yeah, he kept going through cases that could’ve been eliminated, he also kept multiplying by numbers that already made smaller numbers too big for a couple of times before getting rid of them
86 can't be the result of the multiplication because you could only add 12 or 13 which would lead to 98 or 99 making the 8 or the 9 a repeated digit. I think just from the addition 85 would be the highest number that would work, but obviously 85 is not achieveable by multiplication without repeating a 5
I also did it this way, and then brute forced it, starting by getting simple numbers like 8 or 6 in the b's place, which only have one c number
@@Alvin853 you can push it even further, 84 doesn’t work because the only way that you have a five somewhere in the answer is if you add 11 or add 15 (because 5 can’t be in the multiplication), but 11 has a repeating one and adding 15 will give you 99 with a repeating nine, 83 doesn’t work because it’s prime, so 82 is actually the number you should be using as your lowest
@@ezracohen6020 if we're only looking at the addition step, then 83 would be possible with 83 + 12 = 95. But as you said, 83 can't be the result of a multiplication. If we keep going like that' we're doing the brute force in reverse though, 82 can only be achieved by doing 2*41 having a duplicate 2, 81 would need to be 27*3 which leaves no low digits for the addition (lowest digits left would be 4 and 5, 81 + 45 > 100), 80 isn't allowed, 79 would need a single digit number to be added to not have a 9 in the result so that's not allowed, 78 is the first one where you get multiple options
I haven't watched the video yet to see the solution, but my initial thought is "man, that's gonna be a ton of guessing and checking!"
I was thinking he would just do the logic elimination of cases, guess not today 😢
After seeing it was a 22-minute solution, I went to the comments to see if it would be worth while even trying. This comment helps me know not to. Thanks!
You can do a much shorter elimination, leading you to test out only a tiny handful of possibilities, if you focus on the multiplier first; grouping by multiplier sets (multiplier can only be 2, 3 or 4), then by lead digit after that, made it easy enough to eliminate out in my head in a few minutes with only a couple of quick scribbles to check.
I would give you a big and special thanks to you Presh because you helped me think more about how to solve problems like these, something I was unable to do some years ago in high school. I watch your videos since 2017 and I always find myself learning and improving. Thank you!
An outstanding example that sometimes a solution requires work
But
That an organized approach reduces the work greatly!
Aah! and there I was expecting some elegant mathematical solution.
However, It taught me some logical thinking and that sometimes, the only way to solve a problem is to yomp though it. 🤔--->🤯--->🤕--->🤓👍
I looked for how I might solve this more analytically. The following steps were the best I came up with:
1. Prove that if de > 75 then either f = 1 or g = 1
2. Prove 1 < c < 5
3. List the possible combinations of a and c (three where a = 1, three where it isn't)
4. Prove d 5
7. Prove c = 4
8. Complete the solution
A couple of steps (#4 and #7) require individual cases to be completely worked through but only a handful.
The full thing:
1. Prove that if de > 75 then either f = 1 or g = 1
Some combinations of d, e, f and g for de>75 are straightforwardly impossible. If de>88, de>80 and f>1, de>76 and fg>22, or de=76 and fg>23 then de+fg >= 100. If de=76 and fg=23 then hi=99, so h=i. If fg=22, then f=g. fg cannot be 20 as 0 is disallowed. As fg>=12, that means if de>75, fg must be between 12 and 19 or 21: f or g must be 1.
2. Prove 1 < c < 5
If c>6 or c=6 and ab>12, de>75 so f=1 or g=1, so the lowest value of ab must be 23, and ab * c >= 90 so de+fg >= 100. If c=6 and ab=12, de=72, so b = e. If c=5, e would either be 5 (clashing with c) or 0 (disallowed). If c=1, b=e. So 1 < c < 5.
3. List the possible combinations of a and c
a must be less than 10/c else ab * c >= 100. a cannot be the same as c. If c=4 and a=2, then de>80 so f=1 or g=1, so ab >= 23, ab * c >= 90, and de+fg >= 100. Six options remain. If a=1, then c could be 2, 3 or 4. If a is not 1, c=2 and a=3, c=2 and a=4, or c=3 and a=2.
4. Prove d = 100.
If d=8, then f=1 and h=9 so that de+fg 75 means f=1. There are three possibilities of a, b and c giving 75 < de < 80. If ab=38 and c=2 or ab=26 and c=3, then de=76 or de=78 respectively, h would need to be 9 as it must be greater than d and 8 is already taken, so 4 and 5 are left for g and i, which cannot work. If ab=39 and c=2, then de=78 and there is no option left for a value of h > d.
For de = 100. 74 could only be created when c=2 and a=3, meaning ab=37 so b=d. 75 is impossible because e cannot be 5 unless b or c are also.
Therefore, d = 100. So a = 1.
6. Prove b > 5
As a=1, if b * c < 10 (c=2 and b 10 and h = f+d+1, so f must be less than 5, the remaining digits are 2, 5, 7 and 9, so f=2 and h=7, but then g and i would have to be 5 and 9 which doesn't work.
When c=3 and b=8, de=54, the remaining digits are 2, 6, 7 and 9, the only combination which fits e, g and i is g=2 and i=6 but then f and h would have to be 7 and 9 which doesn't work.
When c=3 and b=9, de=57, the remaining digits are 2, 4, 6 and 8, and no combination fits g and i with e.
So c=4.
8. Complete the solution
If b=6 then e=4 which clashes with c. If b=8 or b=9 then d > 6. Therefore, b=7, so d=6, e=8. As h > d and 7 and 8 are taken, h=9. We can fit the remaining values, 2, 5 and 3 to f, g and i to find the solution: 17 * 4 = 68 + 25 = 93.
If the only solution is brute force, then it's not a puzzle.
But... it wasn't brute force.
@Ranoake guy litterally says at 2:43 "let's go through all possibilities "
@@petrkdn8224 but, being smart about it, aka eliminating some of them. It's ... uhm... educated brute force :))
@@petrkdn8224 After literally excluding 99% of the 300K possibilities. He is brute forcing the last step, not the whole problem...
@Ranoake by this logic , hackers aren't bruteforcing passwords either because for example they know a minimum lenght on some website is let's say 7 charachters and it needs to have 3 numbers, so they skip all combinations under 7 charachters and with less than 3 numbers... they eliminate billions of combinations but its still bruteforcing it by checking each possibility...
9:27 that actually jumpscared me so hard XD
Given letters to the squares from a to i from top to bottom:
c1, because times 1 repeats the digits => contradicts different digits (cdd)
c5, because times 5 terminates by 5 => cdd or 0 => not possible
For the same reason, b 5
c9, because then 10a+b cdd
c8, because then 10a+b too much
c7, because then 10a+b too much for addition
13*7=91=>cdd
12*7=84, leaving 35 as the smallest to add => too much
c6, because then 10a+b cdd
13=6=78, smallest to add = 24 => too much
That leaves c = 4 or 3 or 2
c=4 => 10a+bcdd
13*4=52, leaves 67 to add => too much
14*4=>cdd
16*4=64=>cdd
17*4=68
68+23=91=>cdd
68+25=93 => !! SOLUTION !!
68+29=97=>cdd
68+32=100=> too much
18*4=72, leaves 35 to add => too much
19*4=76, leaves 35 to add => too much
21*4=84=>cdd
23=4=92=>cdd
The rest of the reasoning is more of the same somewhat tedious case by case examination showing there's no other solution than the one we found.
d can't be 1( cause a must be 1 too), d can't be 2 ( c and a must be 2 and 1) and d can't be 9 ( because f >=1 , 90+10 >= 100).
And we know, that de = ab*c, so de is not prime. Also we know that de and e5
I think easier to check all variants for de.
That's so much brute-force. First, just obvious check that -ab- < 44, because -de- can't be 9x because otherwise you can't add 2 digits number, brute force for the other 4 numbers to realized that the entire 4x row is invalid, narrow it down to -ab- < 40. Then, b cant be 1 or 5, as you said, and c cant be 1, 5 and 9. Brute force c by putting 8,7,6 on the 1x numbers and those weren't fit too, make the c possibility goes down to 2, 3 or 4. If you made a sheet outside before, there will be 2 -ab- "boxes", the one box that ended in 2, 3, 4, and the other box that ended in 6, 7, 8, 9. You can brute force the 234 box first, it's small enough and the c number will make the elimination faster, and then check the other box, which is just 12 numbers to do manually
I appreciate this channel, this channel is very good. I am from India and learned many ways to solve problems. I usually see video of this channel. This is a good ❤channel❤ for student.
WOW! 362,880 possible ways, I solved it in about 10 minutes on my 6th try! 😁
Not a big deal at all.
I started with the multiplicand and the elimination process is fastet.
I will say a slight shortcut to elimiting answer is that any possible that has d = 9 will not work as d+anything > 9 so there is hard lose cap of de < 90 (you can expand the logic a little more to say it has to be less then 85 because the small number you can therotically ever add is 12 and you can only go up to 97 because of repeat digits)
I am sure there are other cool limitations to the varibles out there that i still have not thought of
yeah, h from get go should be greater than 6, because the smallest a and c are 1 * 2 => smallest d is 3 => smallest f cannot be 1 or 2, it should be at least 3 because for 1 and 2 have already been reserved => h is 6 or greater.
@@The14Some1 love this little observation
Did it pretty much the same way, except I took c as my base for it, since it only has 3 options. 1 is out for repetition, 5 for getting a 5/0, 6+ is out because the smallest option (13*6=78+24=102) is too big. Leaving 2, 3 and 4 as options.
If c=4, A can't be 3+ obviously. So we need to check the remainder (12, 13, 16, 17, 18, 19, 23. Anything bigger overflows). 12 doubles, 16 doubles, 23 doubles, so 13/17/18/19 remain to check the way you did.
If c=3, A can't be 4+, nor can it be 3. So we have to check 12,14,16,17,18,19,24,26,27,28,29. Without doubles, that's 16, 18, 19, 26, 27, 29.
If c=2, then A can't be 2 or 5+, so it must be 1, 3 or 4. So we have to check 13, 14, 16, 17, 18, 19, 34, 36, 37, 38, 39, 43 (46+ makes 92+ so it'd overflow even with +10).
13, 14, 16, 36, 37 double, leaving 17, 18, 19, 34, 38, 39 and 43.
So while it looks daunting, simple elimination only leaves us 4+6+7=17 options to bruteforce check. Which, once you get the hang of it, is pretty fast: options for f and h tend to be limited, if any exist; and g and i can sometimes be eliminated by an odds/evens problem outright. Also note that many of the options to check use the 7. 8 or 9, which eliminates key options for h.
So it takes mainly diligence to check and find 17*4=68; +25=93
My solution did a lot more whittling before going brute force, on only 11 cases.
Define letters as Presh did, so AB x C = DE, DE + FG = HI.
As Presh noted, neither B nor C can be 1 nor 5, as they lead to re-use or a zero in E.
Note that the minimum value for AB and FG are each 12. Likewise, the maximum for GI is 98.
This means that the maximum DE is max(GI) - min(FG) = 86.
As C >= 2, this implies AB =12, DE
I found a few more limitations...like "c can only be 2,3 or 4" and "a can only be 1 or 2". de cannot be prime, end in 5 or 0. This narrows it down to 24 options for de, with 30 corresponding possibilities for ab*c.
It's easy to whittle those down, if you remove ab has to be 2 digits and a,b,c,d,e cannot be repeated. I got 11 such combinations and then brute checked those.
I knew it would be A LOT FASTER to solve this problem by the kind of method used by Presh but I couldn't resist trying to solve it with an Excel spreadsheet. It took "a few" hours to create, but it worked.
Notating top left as A5 and bottom right being B1. a few obvious logic points, A5 has to be 1-4, (likely 1 or 2). B4 has to be 2-4, A1 has to be 6-9 (likely 8 or 9). 5 can't be in the multiplication digits (gives 0 or 5 as a product) etc. Other points- B5 is a nice place to put a big number, as is B3. I don't think I tried more than 10 permutations before I got the answer.
A way to reduce the verifications if the sum is bigger than 100:
Suppose the fifth number is 98 and the fourth number is 12, then the maximum value for the third is 98-12 =86.
Then, we search only three values (a,b,c) such (10a+b)*c
Damn I didn’t see this comment. That’s exactly how I did it. It’s much easier to prove c can’t be high numbers and limit it to 234, and work from there.
This was a horrible problem requiring brute force, I appreciate your patience with "big" numbers because a "xx+12" surprise solution may have existed. I was hoping for a math equation system, but my LIKE goes to the editing quality.
If you are a programmer, this is one case where a little light recursion can save you a whole lot of grief. Nine nested for loops is not a pretty construct. Compare it with this pseudocode:
def test_sum (arguments)
if (we have exactly 9 digits)
then
if (the sum is satisfied)
then print the answer
else just exit
fi
else
for (x = each digit that has not been mentioned)
test_sum arguments, x
next
fi
fed
test_sum 1
In the multiplication section you can eliminate any number over 90, since the addition number must be 2 digits.
This also lets us stop checking at 45x2.
anything over 86, as 98 is the max total (not 100) and 12 is the minimum added (not 10). This lets you stop checking at 43*2 too.
@@richbuckingham true
@@richbuckingham we can probably make it smaller still by arguing that anything over 33 can only be multiplied by 2, thus the addition must be 13 or greater and as you said 98 is the max final value so 85 is the multiplied total, but the multiplication can’t end in a 5, so 84. So we can stop checking at 42x2, which is obviously wrong, so 39x2 would be the last real check.
@@richbuckingham Exactly, that's my first understanding of this puzzle.
I tried the questions before myself, I didn't know the technique he used in this video yet i got the first solution (17 x 4). I was stunned for a moment
You can also do this backwards: picking the highest result number (98), checking for possible subtractions, and eliminating repetitions and indivisible numbers.
Checking most combinations like in the video is a valid way to do this
There are clever eliminations that can be used
1. Noticing we must add another 2digit number, so the product 20 afterward)
check: 4
21,23 don't work. So 12~19*4=48~76, eliminate wrong combinations, left 52 68 72 76
(72+3X>99, 76+2X>85) brute force through 52,68 to find answer 17*4+25=93
check:3
if we use 20s to multiply 3, the equation should be 2A * 3 + 1B = CD (obviously cannot add 4X)
looking at multiplication, only 26*3=78 and 29*3=87 works, checking the addition, no answers here
1A * 3 + BC = DE
multiplication: only 16*3=48, 18*3=54, 19*3=57 works, brute force to find no answers
(Brute force, start from the smallest)
check: 2
30s, 3A * 2 + 1B = CD
34*2=68, 38*2=76, 39*2=78 (68+1X=8Y, fail; 76+14or15 fail; 78+1X
I used a similar start to eliminate 1 and 5 from being B or C. However, I then showed that C could never be greater than 5 because if C is greater than 5, A would have to be 1 (otherwise the multiplication would have a 3-digit result). 12x6=72 and 72+34>100. Similarly, 13x6=78 and 78+24>100, and anything else just results in an even larger 3-digit number after the addition so C could only be 2, 3, or 4. After that, it was easy to prove through brute force that the only solution was 17x4=68 and 68+25=93 similar to how you did.
Another cool trick: Any product that is 88 or above can be eliminated because you have to add a 2 digit number to it that would make the result 3 digits and the smallest possible 2 digit number without repeating is 12
I think you can also eliminate numbers that give you higher than 84 after multiplication. You need to add a two digits number, and the lowest one possible to add is 12. The number after addition can not go above 98, so 98-12=86. But in this case there is a repeated 8, so the number in question goes 1 lower, and we can't have a number with a 5 on the end, so our final number is 84. This could eliminate quite a few possibilities without a need to check them. For example you don't need to multiply by 4 any number in 20-s
I love solving for a solution based on the thumbnail before watching these videos for alternate perspectives. Surprised to see only one for this problem, and error checking isn’t that great
1,7,4,6,8,2,5,9,3 f(from a to i). This has nothing to do with brute force. It's a successive process of elimination.
After investigating combinations with a high likely hood of digits being generate multiple times, and zeros being produced, they were dropped. The main contestants for the initial multiplication (contained in the first 2 fields) are therefore either 17, 18, or 19. The multiplier in the 3rd field can only be 2,3, or 4. 4 bombed out immediately, one case produced a double digit and one went past the boundary of 98.
That left merely 7 combinations with the first 5 digits being in order, plus 4 defined digits to play around with.
Easy as pie. You combine 2 of those leftover digits and add it to what's already known and see if the result matches any combination of the remaining two digits. You only have to do that 7 times! 6 times I failed and on the very last option left being (17x4=68), voila! 68+25=93.
Commenting before seeing the solution: I've brute forced in my head and got 17x4=68+25=93, I wondered if there's any more solutions.
Update after watching: Let's FREAKING go. I'm not joking, I literally worked out the one and only solution, without any papers and pens. That's incredible!!! 😊
Bonus: There's a few more equalities of the same kind that I know on top of my head
54 x 3 = 27 x 6 = 18 x 9
29 x 6 = 58 x 3 = 174
I did much the same. As long as you have the right initial goals in mind it's not too hard to get it quickly. In addition to the stuff with 1 and 5, you know the last 2-digit number is going to be the highest, so you want to try to avoid using any big numbers (6, 7, 8, 9) in your initial multipliers so you can add up to 80- or 90-something at the end, and you want to have a low 10's digit (2, 3) in the second to last number so that the multiplication result doesn't have to be too low. 1 seems like the ideal 10's digit for the multiplication, since with 20 the products start to get big enough that there's no space left before 100, and you don't want to use two odd numbers for the other two because you know you'll probably need the 9 for the last number. Same with using both even numbers, since you're already guaranteed to use at least one more even number for your product if either of your multipliers is even.
All that together, and basically something-teen (odd) times 4 is probably the best place to start. And there are only so many options (basically just 13 and 17). As soon as I saw 68 for the product, before I even considered any actual further calculation, I knew I probably had it, because that's all the large numbers gone but 9 (which I expected to use as the final 10's digit anyway), and 2 still available so I can get to that 90-something sum with ease.
The part where you say any number over 98, I would say cut any number in the multiplication over 85 as the smallest 2 digit number is 12 that also leaves you with a 2 digit number at the end that does not repeat any with 8 is 97. Would help with the multiplication part.
Neat puzzle. After some reasoning and doodling on a scratch pad for 2-3 minutes, I had concluded that c must equal 2, 3, or 4 and that a must equal 1, 2, 3, or 4. From there I did go to a more "brute force / eliminate possible answer" approach and came up with the solution. I had worked in the opposite order as Presh, tackling a = 4 first, then a = 3, et cetera since those have much more restrictions on what c could be. In my opinion, this 22 minute video could have been more like 9-10 minutes, if a bit more time had been spent on the "front end" of eliminating possible digit combinations. As others have said, too much of a "try all combinations" approach ultimately isn't a very satisfying strategy.
I narrowed down the multiplier to be one of 2, 3 or 4. Then I checked the numbers individually with 2, 3 and 4. Starting with 4 to get
17*4=68+25=93
"In a remote and mysterious land, there exists a temple hidden deep within a dense jungle. The temple's entrance is guarded by a series of ancient statues placed at strategic locations.
One particular statue, known as the Guardian, casts a shadow that holds the key to unlocking the temple's secrets. The Guardian's shadow is cast on a sacred stone platform, creating a triangle of shadow.
The temple guardians believe that the shadow triangle is similar to a smaller triangle inscribed on the stone platform. The smaller triangle represents a map leading to the temple's inner chambers.
The lengths of the sides of the shadow triangle are given by the quadratic equation s^2 - 6s + 9 = 0, where s represents the length of one side of the shadow triangle in meters.
Similarly, the lengths of the sides of the smaller triangle are given by the quadratic equation t^2 - 4t + 4 = 0, where t represents the length of one side of the smaller triangle in meters.
To decipher the temple's secrets, the guardians must determine the ratio of the perimeter of the shadow triangle to the perimeter of the smaller triangle. Find the ratio
A) 3:2 B) 4:6 C)1:8 D) 2:5
Find the correct answer
Bro started with a quote and forgot of it, btw idk the answer
the smallest number you can make for the addition is 12, so no value larger than 87 (because 88 is double) from the multiplication has to be checked.
saves some minor time of this bruteforcing
Wow, this has to be one of the most brute force Presh videos I've seen. I was also able to find the unique answer but I had a different approach. I focused on 'c' and I was able to prove easily enough that it had to be 2, 3 or 4. So I would then consider the possible values of a and b for each c, and work through each one. I tried c=4 first and found the solution (and the only one that worked for c=4), but I then proved there were no solutions for c= 3 and c=2. I feel my approach might have been a bit faster but that's maybe up for debate. Still requires a lot of checking.
Normally, I just proceed with the video because I can't solve it. This time I did and I'm glad I did because the video just checked everything instead of skimming the possibilities through logical eliminations
My intuition was that "c", being a multiplier on a two-digit number that must result in a two-digit product, would be the most restricted. So I tried each digit 1-9 as "c", and found that most of the digits led to immediate dead ends. c=1 obviously forced duplicate digits. c=5 forced either a duplicate e=5 or e=0. c≥6 forced three-digit sums at minimum. So "c" must be 2, 3, or 4.
Then, for each c=2, c=3, c=4, I checked all possible "b", and computed "e" modulo 10. This led to only 16 b-c-e triplets with no 0s and no repeated digits: 3x2=6, 4x2=8, 7x2=4, 8x2=6, 9x2=8, 2x3=6, 4x3=2, 6x3=8, 7x3=1, 8x3=4, 9x3=7, 2x4=8, 3x4=2, 7x4=8, 8x4=2, 9x4=6 (all modulo 10).
16 candidates seemed a reasonable amount to brute force, so brute force I did. Each triplet has few, if any, valid values for "a", and "d" can be computed. So it's not much effort to check each triplet. Only the 7x4=8 triplet leads to a solution.
my thought was to create the largest Sum answer you could and then work backwards, since the final answer is the Sum, start with 98 and figure out how to make a sum of 98 with the remaining numbers and work your way down from there.
You could have stopped sooner by keeping in mind that the final step requires a 2-digit non-repeating number which still adds to 98 or less since 99 is repeating. So the final step can't be 1-9/10/11 so the max middle result is 86, assuming 12 for the last step. Although 12-16 would be eliminated too due to repeating 8 for 98, they might work at lower totals. So as soon as 43x2, using all the lowest factors in the first 3 blocks, came out to 86 and the 2 was a repeat, anything past that would have forced an end result above 98. Work it from both ends.
17x4=68 => 68+25=93 :)
I went by testing with the lowest possible values for a and b. I also eliminated impossible solutions like c=5 and b=5 for example.
I did the same general process but I eliminated more options for C before going through the teens. 9-6 got eliminated, because they were out of bounds. The best case scenario would be 13*c+24
For people who want the answer, here it is: 17 x 4 = 68 + 25 = 93
Wrong use of equal sign.
Here is how I found the right answer (I think much more quickly)
I considered the single digit multiplier first. That combined with the two-digit result imposes a major limitation. Second thing to check is the smallest remaining digit to place in the tens digit of the added number.
12 is the smallest multiplier, then 34 is the smallest amount you can add.
9x12 too large
8x12 too large
7x12 too large
6x12 = repeat 2
6x13 = 78 +24 = too large
5 is impossible because of 5/0 ending
4x12 = repeat 4
4x13 = all the small digits used (52, the 10s digit of the added number needs to repeat)
4x16 = repeat 4
4x17 = solution
Brute forcing the combinations would be trivial on a computer, but the algorithm might look cooler if it used a few tricks that humans might apply on paper. Upper bounds on hi (98) and lower bounds on fg (12) to limit the range of de to a maximum of 84, plus some divisibility rules and using sets in combination to manage the remaining options so the whole process looks concise. It was fun on paper.
17*4=68, +25=93
Awesome puzzle ,thank you so much for smart solution from Pearland,Texas USA.
I was surprised you kept going so long. The first thing I did was work backwards and determined the largest possible number at the end is 98. So my equation looks like this:
ab*c+fg=98
Next, the smallest number that fg could be is 12.
ab*c+12=98 -> ab*c=86
The smallest number for c is 2 (yes it is against the rules, but ok for making a bound)
ab*2=86 -> ab = 43
So 43 is the absolute maximum that I could quickly prove that could work. No need to go all the way to 50.
The logic way is to set the smallest digit at the top and largest at the bottom
So we have
1 _
x _
_ _
Wow that was chaos
Instead of checking for 99, you could've eliminated the multiplied values that are greater than 86 itself as a two-digit number of at least 12 is to be added to it to arrive at the final value of less than 99.
Can take it a step further to 85 since 86+12=98, where 8 is repeated and can’t be a solution.
Edit: as another commenter pointed out, it can’t end in 5 so 84 is the highest possible.
@@zxbot4718 It's a minimum of 12 and not exactly 12. But nos 10 and 11 are not of that case, since the former has zero in it and the latter has repeated nos.
@@zxbot4718 And why it can't end in 5. Only the multiple can't end in 5 and not the final sum.
Nice puzzle. But you could have stopped with the maths way earlier. The biggest number in the end could only be 98. And you have do add a 2 digit number to it, which is at least 12... therefore the the difference is at ma 86... This 86 is the maximum for the product (and not 99). - But I understand your solution for this video to keep the numbers simple.
You can ignore all cases resulting over 86. The smallest number you can add is 12. 87 + 12 is 99 which is already too big. So based on that there are lot more cases you didn't have to check
This is brute force, but it’s also a very bad brute force. Although the calculations are minimal, a lot of time is wasted checking digits against 6/7/8/9 for c. As you said, 1 can’t be c, nor 5. But given the sum adds to a 2 digit number, 9 is out since it’s too big, 8 same reason, 7 has a chance with 12 but that’s 84 and you can’t add a 1z number. 6 loses 12 with 72, 13 is 78 but the smallest is 24, and 14 has the same problem as 7x12. Now you’ve eliminated 1,5,6,7,8,9 for c.
4 is fairly simple as the upper limit on it is 4x19, and you can find 4x17 fairly easily. If you’re told this has 1 solution, this is the fastest way to deductively solve. If you need to prove it, you can eliminate the 4s this way, the 3s have an upper limit at 29, and the 2s at 44. 46-49 are needless checks, and any multiplying of 2 or 3 after their limits is unnecessary.
We can start from the digit at C position. Only 2, 3, 4 can go to the C position. Around 10 minutes is enough to get the solution. Thanks
The approach could have been simplified significantly by saying that the multiplication result cannot be higher than 85, because 85 + the smallest allowable two digit number would be above 97 which is the highest allowable result when dealing with the 80s.
I found a sort of analytical approach but with lots of assumptions:
Lets rewrite the question to
ABxC=DE + FG=HI
First, looking at the one's digit, there are 6 possible cases:
1. even×even=even +even=even
2. even×even=even +odd=odd
3. odd×even=even +even=even
4. odd×even=even +odd=odd
5. evenxodd=even +even=even
6. evenxodd=even +odd=odd
Second, we can note that there are more odds(1,3,5,7,9) than even(2,4,6,8).
Therefore, we can compare each cases and assume we want the one with the most odds/least even.
(Cases 4 and 6)
Another assumption is that we'll start with a small number and end with a large number. Thus, we will assume the A=1 and H=9.
After all these assumptions you can reach the answer by yourself.
The lines after this is for those who are stuck or want to continue the solution:
We can note all the locations of the odd numbers (A,B or C,G,H,I) and determine the location for even (B or C,D,E,F).
Next, since D and F are both even. The only way they can be 9 is if D+F=8 and then E+G>10
So, D and F are either 2 or 6
E is even, so it is either 4 or 8
Cases for E+G>10:
If E is 4: G is 7
but 4+7=11 and 1 is already used
This means E is 8:
8+3=11 (can't be because 1 is already A)
8+5=13 which is possible
8+7=15 which is possible
In both possibles cases 5 is being used.
Looking at the one's digit:
BxC=E(8)
Options remaining are 3,4,7
Only possible cases are 4×7 or 7×4
Therefore, I=3 and G=5
Since B and C are either 4 or 7:
D cannot be 2 and must be 6, F=2.
Thus ABxC=DE is 1(4 or7)×(7 or 4)=68
Which makes B=7 and C=4
Again, lots of assumptions but a more funner way than brute force.
Man, I'm not sure if 20 minutes of brute force calculations is why I watch this channel....😊
instead of just eliminating results over 99, you can immediately eliminate any result where the multiplication is greater than 89. There is no 2 digit number possible to add to 90 to get back a 2 digit number.
Dude! That was awesome.
Im really proud i solved this problem on my own lol
Through setting up a main logic grid, and then a smaller one once your C possibilities are dramatically reduced, you can narrow it down to about 12 possible answers for the "DE" product. Then just check those with the remaining digits. You checked so many unnecessary cases.
My thought process:
maximum final result is 98, since its 2 digits and cannot repeat the 9.
d must be less than 9 so any multiplication over 90 can be discarded
c cannot be 1 as that means a=d and b=e. and b cannot be 1 as that would make c=e
this limits a to
I only checked for C term and started with the largest possibility and slowly went down
I reached from 9,8,7,6,5 and then when I checked for 4, it worked out. I started checking with 12, 13, 17 and found the answer
I found this solution too but was hoping there was a more elegant solution than check all viable options. Guess not. I did eliminate intermediate values at 86 though instead of 99 since the minimum number you can add is 12 and the max valid result is 98.
use a little logic to try to figure out the top, and it was a bit bruteforce at the end but got there. I mostly was like "top has to be small, bottom probably has a 9 as a target, and b&c cant be 1&5"
This problem might have 2 interpretations. 1. This could be a 2 step multiplication algorithm with the final answer on the bottom. 2. It could be interpreted like in this video
You mean long multiplication? Wouldn't that necessarily give a 3-digit result?
Exactly why I was confused. It's poorly presented.
I kinda worked backwards a little.
I knew, based on the criteria, that the final solution could be no larger than 98. In order to get a final solution that was no larger than 98, the multiplication solution had to be no larger than 76. (I knew this because 98, the largest possible double digit combo, minus 12, the smallest possible double digit combo, is 86. But 86 would repeat the 8, as would anything down to 80. 79 would repeat the 9, 78 would repeat the 8, and 77 would repeat the 7.)
This meant that my multiplier (the one single digit number) can't be anything higher than a 6. Again, the smallest possible double digit combo is 12, and 12 times 6 is 72; any larger multiplier (be it 13 times 6 or 12 times 7) would result in a multiplication solution higher than 76. But 12 times 6 isn't right, because 72 would repeat the 2. I also knew, for the same reasons the video stated, that the multiplier couldn't be either 1 or 5. That eliminated my multiplier to either 2, 3, or 4.
So I started with the multiplier 4, as that would have the fewest multiplications I would have to check. Keeping in mind that my multiplication solution can't be any higher than 76, I divided 76 by 4 and found that it's a clean 19. But that wouldn't work because the smallest double digit addition possibility was 23, and 23 plus 76 was 99, which would not only repeat the 9 but also be higher than the final solution of 98.
So I moved on to 4 times 18, which is 72. But that wasn't right because the smallest double digit addition possibility was 35, and 35 plus 72 is 107, which is higher than 98.
Then I moved on to 4 times 17, which is 68. The smallest double digit addition possibility was 23, and 23 plus 68 was 91, which repeated the 1. The next double digit addition possibility was 25, and 25 plus 68 was 93. And this was the correct answer.
I didn't time myself on this process, but it didn't take as long as this video did.
Beautiful reasoning!
You keep eliminating anything over 99 but you could use anything over 88 - the smallest 2 digit number is 12. In practice, for the cases where you are using 1, the smallest 2 digit possible is 23 and you could eliminate anything over 77.
I don't want to be petty, but I see a better way for constraining the numbers are possible.
As 0 is not allowed if the number (of the two added) is greater than 90, you will have to add a number less than 10.
So that is not allowed.
This (to me) is a better filter for the resultant addition.
34*2=68+19=87 too easy
Close but you used 8 twice and didn't use 5.
I assumed the first digit would be a 1 or a 2 and the third would also be low. Guessed it very quickly making similar assumptions, like at least one of the 8 or 9 would be in a singles column in the addition section and the solution would likely start with 8 or 9. Wanting to save low digits for later, guessed 17 in the first spot. After stupidly trying 17 x 3 and seeing I already had a repeat, tried 17 x 4. With a last digit of 8, saw the 5 in the digit column of the next number would give the 3 and the rest fell into place. Bottom line is that I got lucky and got it with just that first 17x3 error.^.
no need to check after 43.. because plus number is more than 12 if you multiply 46x2 = 92.. then you need to only 7 to get 99... so it is not possible as you need to add 07 and zero is not an option..
"There are 9! = 362 880 possible ways"... proceeds to check all 362 880 ways.
Well that was disappointing. I was just trying to solve it from the thumbnail and used logic to eliminate several possibilities, but was still left with a lot of options. I figured there had to be some other logical deduction I was missing, but it turns out this was just a 'run through the remaining options until you find something that works' problem. More of a tedious waste of time than a fun puzzle. It's cool that it only had one solution, but extremely uninteresting in how one must find that solution.
This is much shorter to brute force::
first thing: *c cant be 1* ,so c is 2 or higher, thus *a cant be 9,8,7,6,5* which will result in 3digit number. Also *b cant be 5* as if it was-then e will be 0 or another 5 which both cant happen, *c cant be 5 as well* if it was-then e will again be 0 or another 5 which both cant happen. *e cant be 5* if it was,then either b or c should be 5.
Now with this thing: go through every possibility of e from 1 to 9(5 cant come)
the above conditions such as *a cant be 9,8,7,6,5* eliminates many possibility.
We end up with e=8, i.e. 17*4=68, now with remaining 2,5,3,9 we know f is 2 or 3 else it will give 3 digit sum,and we discover that it is 17*4=68,68+25=93
Python solves it in a fraction of a second :)
from itertools import permutations
for digits in permutations(range(1,10)):
if (digits[0]*10 + digits[1]) * digits[2] == digits[3]*10 + digits[4] \
and digits[3]*10 + digits[4] + digits[5]*10 + digits[6] == digits[7]*10 + digits[8]:
result = digits
break
print(result)
Really amazing
Could have stopped at the 90s because after you have to add a 2 digit number. 90 + any 2 digit whole number will be > 99
Love from India ❤
Same 😂
Your references to "99" should be references to 98, because 99 already contains a repeated "9".
The initial multiplication cannot add up to more than 86, because the minimum later addition is 12.
That means no initial number above 43 works. That also rules out 43 as 2 would be repeated, so much of the tail end analysis could have been omitted.
It is one of the downsides of Math: too mcuh work for too little results!
You can eliminate number > 87 in the multiplication because the final result is a 2-digit number and zero (0) is not allowed.
Thank you. I like this puzzles. Just a little comment: There can be elimanted all solutions over 98, because 99 have two same digits.
You could have restricted the values of the addition part significantly. The "DE"
If you make the puzzle complete, you include the digit 0, which has to be placed in the second row where one square is missing.
Can we get Wolfram to solve it? tried to tell it:
set {1,2,3,4,5,6,7,8,9} = {a,b,c,d,e,f,g,h,i}; 10ac + bc = 10d+e; 10d + e + 10f + g = 10h + i
but it just asigned the a-i in order :-(
Someone may be able to help here. Certainly it can be solved with a few lines of code, though I found it very rewarding to work out by hand :)
Yes, tried by hand first, got a few "that number can't go there", but run out of ides, and didn't want to brute-force it by hand, so did it in php. then watched your video, and noticed you to kind of brute-forced it in the end. Then I started to wonder how to solve it in wolfram.
How are you able to send this 5 days ago?
While the video is like 45 minutes ago
I played around with it too. It's hard to say what syntax wolfram alpha will take to understand the question. It didn't seem to like the set equality statement in general.
accidentally solved it while trying to eliminate 4 from the single digit multiplier, and 17 was the first possibility where i had something going