I was doing this by myself and FINISHED IT, but then I thought there was a fault in the problem because I got two answers. BUT. When Brian said they only got one of two... I felt like a genius.
I actual like your sponsor portions of your videos because they are always new and interesting. (I also love math games. I may not be good at them, but I like them.)
If a puzzle has multiple correct answers, that you get in a very obvious way... not a good puzzle. Or, to put it another way, this wasn't a puzzle. Just a basic equation you'd see in middle school. I'm glad your stuff got better over time ^_^ can actually see you evolve from a "meh" channel to something that puts everyone else to shame. Kudos.
I'm always checking previous videos and current ones for stuff I haven't watched yet. I love scam school. I think Brian should collaborate with other cons and magicians like he does with Diamond Jim. That guy is great too.
I wish they would have said that there were two correct answers at the START of the video. I paused it and wasted a whole lot of time trying to figure out why I couldn't find a unique solution.
This was indeed not cool. Even a 'there might be more than one answer' or 'find all possible answers for a beer' but some way to let us know or suspect that there is more than a unique answer would be nice. But then again Brian also got in the way of the crew working the puzzle a number of times in his excitement. Not a great episode. But thank you for your comment, because i scrolled down to see if someone discussed this exact issue.
I know I'm bragging here, but I found both answers in less than two minutes after watching just the intro. Aside from the bragging, I would like to say that consistently watching Scam School is one of the reasons I was able to do this.
paused the video and did the puzzle myself and ended up scratching my head because I had two possible solutions. Watched the rest of the video and found out they are both correct and now I feel like a genius :D.
If yo take the remaining numbers that don't get used in either solution you get 2, 4, and 9. if you turn those into letters with a simple number-to letter cipher you get B, D, and I. If you rearrange the letters it can spell "Bid"! So if you want have fun with in you can present it with: "Solve this puzzle and you will see; what remains will be the key; a proper welcome from one to thee." And with a glorious red herring in the 123 pun at the end.
I was mad and thought I got it wrong because there were two answers. It was messing with me and I didn't wanna spoil it. Glad I finally gave up to see I was right!
Dude, I paused for a while to try and figure it out, I was stumped because I got down to 2 valid answers (655 and 855) and I played the video to find a mistake, but yall confirmed I was right!
Saw that E had to be one, figured D was probably five making N zero. Then just used six as my first option for O to try it out and it was right. Took about 30 seconds to get the solution, but I didn't think to check for any other solutions.
E has to be 1, O has to be grater than 5. D has to be 5 which forces N to be 0. I'm pretty sure those are the only requirements. For example: 555 + 555 = 1110; 655 + 655 = 1310; 755 + 755 = 1510; 855 + 855 = 1710; 955 + 955 = 1910.
before solution: o must be 5-9 (Brian's hint) e must be 1 (three digit number plus three digit number can't be more than 2000) d must be 5 (third digit in EVEN must be one, so dd + dd must leave a 1 in the tens place) n must be 0 (55 + 55) if o is 5, 6, 7, 8, or 9, v will be 1, 3, 5, 7, or 9 respectively after number can't repeat rule: 655 + 655 = 1310 or 855 + 855 = 1710 after solution: lit, got it right
I honestly just stumbled across the right answer when I was going through the options for D. Assuming each letter had to be a different value, and the fact 5 worked to solve for E, I just moved O to 6.. o.o
One thing he didn't mention in the video is that these equations are called "Cryptarithmetics". I see some of you in the comments found this easy. if you want a challange, try this one: SODAS - AIM = BEER @Brian This video went excelently! I'm so glad you got the sudoku aspect of the puzzle! This is how you can tell some tru puzzles apart from other... it's also why Minesweeper can be a brilliant scam of a puzzle.
Since ODD+ODD is 5 digits, O>4 and E = 1 Since DD + DD = 1N or 11N, D = 5 and so N=0. 2(O55) = 1V10, O can be 5,6,7,8, or 9, and V will be (2O-10)+1 = 2O-9. If numbers can't be repeated, then O isn't 5, it's also not 9 since then V=9, and it's not 7 since that makes V=5. Thus, O = 6 or 8, and V = 3 or 7, respectively. ODD+ODD=EVEN 655+655=1310 855+855=1710
As an algebra lover, this was painful to watch. One does not simply try out the numbers, it's much simpler and more professional to just write down equations (e.g. 2D=N and 2D=E or 2D=N+10 and 2D+1=E) etc.. Eventually solving em all. By the way for anyone interested there's an app called cryptogram that gives a few of these problems, I think they're really fun.
I was getting mad becasue i thought there was only one answer and I couldn't logically figure out why both set of numbers couldn't work...then he said there were 2 answers lol
hey Brian! is there any chance you could do a compilation of tricks with pens? I thought about this because I'm in high school and once in class I was thinking "wouldn't it be cool if I were just vanishing a pen repeatedly in the back of the class until someone just turned around and said :how the hell are you doing that?!"
Paused the video before the two minute mark and in about a minute got the values. If you haven't figured it out, don't keep reading, but O = 6, D = 5, E = 1, V = 3, and N = 0. Educated guesses FTW!
You can also solve it graphically, although its a bit mathy... First do some algebra and rewrite the problem: 2(ODD)=EVEN 2(100*o + 10d+d)=1000e+100v+10e+n 200o+22d=1010e+100v+n Next we know that the one's place of d+d has to equal n because it is the last letter, we can write this algebraically as follows: mod(2d,10)=n We also know that e has to be 1 based off of the logic explained in the video So plugging these into the above equation we get: 200o+22d+1010 + 100v + mod(2d,10) We now have three variables but we know that o can only be 5,6,7,8,9 and so we can write an equation for each of these values Now we have 5 equations which we can graph as a function of v For simplicity I have made the following substitutions: v=y, d=x This gives us: y = (200*o +22d - 1,010)/100 Graphing this, as I did here: www.desmos.com/calculator/dreddhgqse allows us to visually see that the only integer solutions to the equation required x to be 5 when x (which is d) is 5 in each of our six equations we see that y (which is v) must be 1,3,5,7,9 when o is 5,6,7,8,9 respectively We now know that these are the only possible answers and they give us the following 555 + 555 = 1110 655 + 655 = 1310 755 + 755 = 1510 855 + 855 = 1710 955 + 955 = 1910 Finally, we can sort through these to see that they satisfy o=/d=/e=/v=/n and we are left with 655 + 655 = 1310 855 + 855 = 1710 And, well, this is exactly what they found in the video... but at least we have a proof for it now :p
I had it before they started working it out but then you said to focus on odd and even so i started thinking it was wrong because 6 and 8 weren't odd...i still got it in about a minute and a half so I'm happy anyway xD
If we force rule that different letters correspond to different digits, then we are left only with: 655 + 655 = 1310, 855 + 855 = 1710 It's much easier to solve if you write it like it's long addition: _ODD _ODD EVEN you instantly get E = 1, so it is _ODD _ODD 1V1N and here we have not much left for letter D: two of them have to sum up to odd number 1, so there has to be carry out from previous position, and it was D as well, so D is at least 5. Also we have to have exactly 1 so without carry it has to be 0, so D is either 0 or 5, but previously we deduced that it is at least 5 therefore it's 5, so now we have: _o55 _o55 1V1N after that we get that N = 0 and for o we've got only 4 different values we need to check, 6, 7, 8 and 9, from which only 6 and 8 satisfy condition that all digits have to be different. Lol, this is pretty much majority of things which children do at 4th-5th grade in my country and this is quiet useful for developing logic and ability to deduce things, I don't understand why it took adults so much time, it's very easy.
1:44 Oh, cool, like Sideways Arithmetic from Wayside School! (In fact, they probably used this one, but I've since forgotten it.) Okay, let's figure this out... Two digits added together will never be greater than 18, so if you carry anything, it will be a 1. ..ODD +ODD _____ EVEN With that said, there was nothing in the thousands column, but we ended up with E. So we know E is 1: ..ODD +ODD _____ 1V1N Now, looking at the tens column, we know D+D should be even, but there's a 1 under that sum. So there must be a 1 carried over from the ones column, and D+D ends in 0. But the units column, which we just said must carry a 1, is D+D. Since D+D both ends in 0 and carries a 1, it must be 5+5=10 (which also means that N must be 0). ..O55 +O55 _____ 1V10 Now we know there's a 1 carried over to the 100s column, (and that the result of the sum "carries" over to the 1000s column), so O+O+1=V+10. Rearranging a little, 2*O=V+9. Now if V were even, V+9 would be odd, but it's obvious that 2*O is even, so V must be odd. (And so 2*O=V+9 an ODD no. plus an ODD no. equalling an EVEN no., reminiscent of the problem itself!) I assume the digits are different (other comments seem to confirm that he eventually says this), so V can't be 1 or 5. That leaves 3, 7, and 9. We'll have to test these through brute force. Solving for O, we know O=(V+9)/2. That should give us a good starting point. And the key to eliminating possibilities is that all letters are different digits, so if one letter (in particular, O) is the same as another, we have a contradiction and our premise must be wrong. If V=3, O=(3+9)/2=6 If V=7, O=(7+9)/2=8. That could be. If V=9, O=(9+9)/2=9. But then O would be the same as V, so that can't be. So I can't work it out right now, but I've only started the video. Maybe there's a rule that none of the letters are in the names of the digits they represent? Then it would have to be ..855 +855 _____ 1710 but I haven't heard all the rules yet
I just looked at the d and said ok what double digit will set me up like this at the end. idk I'm good at math but I actually did 55+55 first and I figured out there was actually multiple answers before I even finished the puzzle but I chose 655.
1. E must be 1 because two 3-digit numbers added together to create a 4-digit number must be at least 1000, but _cannot_ be 2000. 2. (Since that makes the 3rd digit of EVEN also 1), DD + DD must be 55 + 55 because it is the only combination of two 2-digit same-digit numbers (11, 22, 33, 44, 55, 66, 77, 88, 99) added together that results in a 1 in the tens column. 3. Therefore, N must be 0, since 55 + 55 = 110. 4. O + O + 1 (carried from DD + DD) is at least 10 (as EVEN is 4 digits), so O must be at least 5. There is not enough information to solve, but possible solutions that do not repeat any digits from other solutions are 6 and 8. (655 + 655 = 1310, 855 + 855 = 1710). 555, 755, and 955 do not work because they repeat digits: 555 + 555 = 1110 (that obviously repeats a few), 755 + 755 = 1510 (D and V would both be 5), and 955 + 955 = 1910 (O and V are both 9).
So for those doing this at home and are reading the comments before watching the full video, two important facts that weren't mentioned initially: 1) Each letter represents a _unique_ integer. 2) There are two solutions. (If the letters don't represent unique integers, there are four solutions).
he doesn't have to state that each letter is a unique integer as that is very obvious else there would've been hundreds of solutions such as 666+666= 1332, 667+667= 1334 etc
posting before answer is shown but i got it figured out O=Any # > 4, D=5, E=1 N=0 ,and V is a little tricky but first i will create a new variable x which is for solving for v, x = 9 - O, And v = O - X
Says, "algebra question", then has lets represent digits with implied concatenation? Seems to me, it should be multiplication if you wanted to claim it is an algebraic statement.
The equation for this problem is E times the square root of 3 plus V times the square root of 3 plus N times O = 2 times D plus 2 times D. This means only 855+855=1710 can be the answer. The other, 655 is a false positive.
Problem: Find all combinations of digits that satisfy this pattern, each number must be different: ODD + ODD = EVEN O=6,8 D=5,5 E=1,1 V=3,7 N=0,0 Or: 655 + 655 = 1310 855 + 855 = 1710 I believe these are all the solutions. Logic: First e must equal 1, because it's two three digit numbers, and there is no way for three two digit numbers to be added to produce a number greater than or equal to 2000. (999+999 = 1998, the highest we can go). Therefore, the target number is bounded between 1000 and 1998. For O since the answer is bounded between 1000 and 1998, we know O must be greater than or equal to 5, and less than or equal to 9. That's all we know for now, we'll come back. For D, we need to produce the 1v1n pattern, so the 10s place must have a 1 in it. The selected 0 doesn't affect the 10s place, it only affects the 100s and1000s place (we can convince ourselves by this line of reasoning: ODD + ODD = 2*ODD = 2*(O*100 + D*10+D) = (200*O + 20*D+2*D), so we can clearly see for the 10s place and 1s place, we see that O will be multiplied by 0. and will have no effect, but D will. . That means DD*2 must have a 1 in the 10s place. We can try all 10 numbers if we want (00,11,22,33,44,55,66,77,88,99) to convince ourselves that 110 is the only number that will produce this, so D *must* equal 5, since 55*55 = 110. Okay, now that we've got that sorted, we know N has to be 0. Then V depends on our selected O. Again, we try all 5 values of O (5,6,7,8,9) and we find that the only ones that have a digit in the V place that isn't already taken are 6 and 8. Edit: I just saw their solution part of the video, just to explain why O=9, O=7, and O=5 don't work (for the sake of completeness of my own solution), it is because the digits are already used. D is already 5, so O can't be that, O=7 doesn't work, because it gives 1510, but then V=D=5, which isn't allowed, and O=9 doesn't work because it gives 1910, so then O=V=9, which isn't allowed because they say you must have a different number of each letter. :)
For those interested in a mathematical proof of the solutions, I came up with this (feel free to correct me): Let's find the digits O,D,E,V,N making ODD + ODD = EVEN Because the two same 3-digits numbers sum up to a 4-digits number, we know that O ≥ 5. We also know that two 3-digits numbers can't sum up to more than 1998 (in general x-digits number summing up to a (x+1)-digit number will be 1-something). Which means that E = 1. Let's suppose that D < 5. Then D + D = N D + D = E But that's impossible, because all digits are different. Which means that D ≥ 5. This means that D + D = 10 + N (N is the last digit of a number between 10 and 19) E = (N + 1)%10 (that's D+D plus the 10 of the previous one) EV = O + O + 1 So far we have (solving the equations above) E = 1 N = 0 D = 5 This leaves us with O and V. We know that O ∈ {6,7,8,9}. If we add 1 to the double of each of these numbers, we get V. So V ∈ {3,5,7,9}. But each digit is unique, which leaves us with (O, V) ∈ {(6, 3), (8, 7)} Which leaves us with two only possible solutions 655 + 655 = 1310 855 + 855 = 1710 Please do correct!
![ลองสุ่มไอโฟน 990 โกงมั้ย? [ โกงมั้ยครับ ep.101 ] | DOM](http://i.ytimg.com/vi/bfsWe0j492k/mqdefault.jpg)
I was doing this by myself and FINISHED IT, but then I thought there was a fault in the problem because I got two answers. BUT. When Brian said they only got one of two... I felt like a genius.
"I'm not helping you guys anymore!"
"Let me just give you one more piece of advice..."
Lol Brian.
"The V can't be the same as the D"
- Brian Brushwood 2017
Try telling that to an activist
@@spicytrashpanda damn I forgot I said this 😭😭😭 I'm queer now 🏳️⚧️🏳️🌈
Once you have eliminated the impossible whatever remains, however improbable, must be the truth.
That bald signal is just the moon.
there's actually a whole book with stuff like this, Sideways Arithmetic from Wayside School by Louis Sachar
Those domain.com ads are so good I think I'll buy a domain anyway even though I don't need one
Before Brian said that they were all different numbers, I though this was a really bad puzzle... 000+000=0000
Tommy Dancer Zero isn't odd.
Kyle Kern Did he state that the number had to be odd? (Also zero's not even either) (Never mind I got to the point in the video he said it)
Zero is even. It fits all the criteria.
And that's numberwang!
Tommy Dancer 59+41=100
655 + 655 = 1310
I just literally started by plugging in a 5 for D, saw it fit, plugged in a 6 for O, saw that fit, and suddenly was done. Oops.
I actual like your sponsor portions of your videos because they are always new and interesting. (I also love math games. I may not be good at them, but I like them.)
If a puzzle has multiple correct answers, that you get in a very obvious way... not a good puzzle.
Or, to put it another way, this wasn't a puzzle. Just a basic equation you'd see in middle school.
I'm glad your stuff got better over time ^_^ can actually see you evolve from a "meh" channel to something that puts everyone else to shame. Kudos.
I'm always checking previous videos and current ones for stuff I haven't watched yet. I love scam school. I think Brian should collaborate with other cons and magicians like he does with Diamond Jim. That guy is great too.
I wish they would have said that there were two correct answers at the START of the video. I paused it and wasted a whole lot of time trying to figure out why I couldn't find a unique solution.
This was indeed not cool. Even a 'there might be more than one answer' or 'find all possible answers for a beer' but some way to let us know or suspect that there is more than a unique answer would be nice. But then again Brian also got in the way of the crew working the puzzle a number of times in his excitement. Not a great episode.
But thank you for your comment, because i scrolled down to see if someone discussed this exact issue.
Glad it wasnt just me!
I have to admit. That ad in the video was so smooth.
I figured out the answer insanely quick on this one. I love that type of puzzle and I've done quite a few of them. I even made them at one point.
Really good puzzlers used here. Great vid.
I know I'm bragging here, but I found both answers in less than two minutes after watching just the intro. Aside from the bragging, I would like to say that consistently watching Scam School is one of the reasons I was able to do this.
Did it without using pen/paper. Many of your puzzles are too easy man, some are insanely awesome.
Wow! first time I won ahead of the others. I just went at it differently but also came out with the same answer option. Loved it!
Loved the ad read
I spent about 10-15 minutes on this when he gave the prompt. Was super happy with myself when I got it.
Oh. I missed something that would have made it easier and faster. Eh. Still got it.
paused the video and did the puzzle myself and ended up scratching my head because I had two possible solutions. Watched the rest of the video and found out they are both correct and now I feel like a genius :D.
655 and 855 work, but 7 and 9 repeat. Oh, you got it as I wrote this. As you were!
If yo take the remaining numbers that don't get used in either solution you get 2, 4, and 9. if you turn those into letters with a simple number-to letter cipher you get B, D, and I. If you rearrange the letters it can spell "Bid"! So if you want have fun with in you can present it with: "Solve this puzzle and you will see; what remains will be the key; a proper welcome from one to thee." And with a glorious red herring in the 123 pun at the end.
King Kasper very clever
Gave up after finding the right answers because I thought there could only be one answer. Gdi Brian.
'You saw the bald signal'
...That's the moon, Brian.
I was mad and thought I got it wrong because there were two answers. It was messing with me and I didn't wanna spoil it. Glad I finally gave up to see I was right!
like a $0.25 paperback mystery novel, you changed the rules at the very end.
Trevor is finally in an ad! \o/
Trevor is great!
More than that, he's finally back
this was insanely easy I knew the answer right away
"You saw the bald signal..."
The bald signal? You mean just a regular signal light? 😂😂😂
I sat there a good fifteen minutes thinking that their couldn't be two right answers. Then gave up and watched the rest, guess I was right!
bald guy with glasses is the mvp of this group.
Dude, I paused for a while to try and figure it out, I was stumped because I got down to 2 valid answers (655 and 855) and I played the video to find a mistake, but yall confirmed I was right!
Finally all of the math I learned in High school and collage paid off :p
Brian was a proud parent in this episode.
Solved it in my head in a couple of minutes. Once I realized E was 1, it was quick to get D was 5 and then it was easy.
Whoever the video needs a raise!
As soon as Schwood said algebra I thought: (O*D*D) + (O*D*D) = E*V*E*N. Glad it wasn't that complicated.
Saw that E had to be one, figured D was probably five making N zero. Then just used six as my first option for O to try it out and it was right. Took about 30 seconds to get the solution, but I didn't think to check for any other solutions.
E has to be 1, O has to be grater than 5. D has to be 5 which forces N to be 0. I'm pretty sure those are the only requirements. For example: 555 + 555 = 1110; 655 + 655 = 1310; 755 + 755 = 1510; 855 + 855 = 1710; 955 + 955 = 1910.
I solved this in like five mins. this is one of your easier puzzles thank you.
before solution:
o must be 5-9 (Brian's hint)
e must be 1 (three digit number plus three digit number can't be more than 2000)
d must be 5 (third digit in EVEN must be one, so dd + dd must leave a 1 in the tens place)
n must be 0 (55 + 55)
if o is 5, 6, 7, 8, or 9, v will be 1, 3, 5, 7, or 9 respectively
after number can't repeat rule:
655 + 655 = 1310 or 855 + 855 = 1710
after solution:
lit, got it right
I honestly just stumbled across the right answer when I was going through the options for D. Assuming each letter had to be a different value, and the fact 5 worked to solve for E, I just moved O to 6.. o.o
love this show
Do the 12 balls, 3 weighing logic puzzles. It would probably be the hardest puzzle you have done on this channel.
Trevor with the STL Cardinals hat 🙌🏻
Either I'm getting to smart or the puzzles are getting to simple for me to think there actually are puzzles
Hah - "Hope y'all got degrees in Russian finance" - Yeah, sage advice it seems ;)
One thing he didn't mention in the video is that these equations are called "Cryptarithmetics". I see some of you in the comments found this easy. if you want a challange, try this one: SODAS - AIM = BEER
@Brian This video went excelently! I'm so glad you got the sudoku aspect of the puzzle! This is how you can tell some tru puzzles apart from other... it's also why Minesweeper can be a brilliant scam of a puzzle.
here are all 48 answers to the soda beer
#1
sodas: 2450
aim : 563
beer : 1887
~~~~~~~~~~~~~~~
#2
sodas: 2450
aim : 567
beer : 1883
~~~~~~~~~~~~~~~
#3
sodas: 3450
aim : 561
beer : 2889
~~~~~~~~~~~~~~~
#4
sodas: 3450
aim : 569
beer : 2881
~~~~~~~~~~~~~~~
#5
sodas: 3460
aim : 681
beer : 2779
~~~~~~~~~~~~~~~
#6
sodas: 3460
aim : 689
beer : 2771
~~~~~~~~~~~~~~~
#7
sodas: 3560
aim : 671
beer : 2889
~~~~~~~~~~~~~~~
#8
sodas: 3560
aim : 679
beer : 2881
~~~~~~~~~~~~~~~
#9
sodas: 4250
aim : 581
beer : 3669
~~~~~~~~~~~~~~~
#10
sodas: 4250
aim : 589
beer : 3661
~~~~~~~~~~~~~~~
#11
sodas: 4560
aim : 671
beer : 3889
~~~~~~~~~~~~~~~
#12
sodas: 4560
aim : 679
beer : 3881
~~~~~~~~~~~~~~~
#13
sodas: 4690
aim : 912
beer : 3778
~~~~~~~~~~~~~~~
#14
sodas: 4690
aim : 918
beer : 3772
~~~~~~~~~~~~~~~
#15
sodas: 5190
aim : 963
beer : 4227
~~~~~~~~~~~~~~~
#16
sodas: 5190
aim : 967
beer : 4223
~~~~~~~~~~~~~~~
#17
sodas: 5690
aim : 912
beer : 4778
~~~~~~~~~~~~~~~
#18
sodas: 5690
aim : 918
beer : 4772
~~~~~~~~~~~~~~~
#19
sodas: 6230
aim : 341
beer : 5889
~~~~~~~~~~~~~~~
#20
sodas: 6230
aim : 349
beer : 5881
~~~~~~~~~~~~~~~
#21
sodas: 6280
aim : 831
beer : 5449
~~~~~~~~~~~~~~~
#22
sodas: 6280
aim : 839
beer : 5441
~~~~~~~~~~~~~~~
#23
sodas: 7230
aim : 341
beer : 6889
~~~~~~~~~~~~~~~
#24
sodas: 7230
aim : 349
beer : 6881
~~~~~~~~~~~~~~~
#25
sodas: 7340
aim : 451
beer : 6889
~~~~~~~~~~~~~~~
#26
sodas: 7340
aim : 459
beer : 6881
~~~~~~~~~~~~~~~
#27
sodas: 7280
aim : 831
beer : 6449
~~~~~~~~~~~~~~~
#28
sodas: 7280
aim : 839
beer : 6441
~~~~~~~~~~~~~~~
#29
sodas: 7380
aim : 821
beer : 6559
~~~~~~~~~~~~~~~
#30
sodas: 7380
aim : 829
beer : 6551
~~~~~~~~~~~~~~~
#31
sodas: 7490
aim : 932
beer : 6558
~~~~~~~~~~~~~~~
#32
sodas: 7490
aim : 938
beer : 6552
~~~~~~~~~~~~~~~
#33
sodas: 8290
aim : 954
beer : 7336
~~~~~~~~~~~~~~~
#34
sodas: 8290
aim : 956
beer : 7334
~~~~~~~~~~~~~~~
#35
sodas: 9130
aim : 354
beer : 8776
~~~~~~~~~~~~~~~
#36
sodas: 9130
aim : 356
beer : 8774
~~~~~~~~~~~~~~~
#37
sodas: 9270
aim : 714
beer : 8556
~~~~~~~~~~~~~~~
#38
sodas: 9270
aim : 716
beer : 8554
~~~~~~~~~~~~~~~
#39
sodas: 10231
aim : 345
beer : 9886
~~~~~~~~~~~~~~~
#40
sodas: 10231
aim : 346
beer : 9885
~~~~~~~~~~~~~~~
#41
sodas: 10241
aim : 463
beer : 9778
~~~~~~~~~~~~~~~
#42
sodas: 10241
aim : 468
beer : 9773
~~~~~~~~~~~~~~~
#43
sodas: 10251
aim : 584
beer : 9667
~~~~~~~~~~~~~~~
#44
sodas: 10251
aim : 587
beer : 9664
~~~~~~~~~~~~~~~
#45
sodas: 10281
aim : 835
beer : 9446
~~~~~~~~~~~~~~~
#46
sodas: 10281
aim : 836
beer : 9445
~~~~~~~~~~~~~~~
#47
sodas: 10381
aim : 824
beer : 9557
~~~~~~~~~~~~~~~
#48
sodas: 10381
aim : 827
beer : 9554
~~~~~~~~~~~~~~~
I could see this trick being effective with drunk people.
Paused and got it in 6 minutes.
But I was sober, so I guess I didn't do any better than Team Melon Head.
Since ODD+ODD is 5 digits, O>4 and E = 1 Since DD + DD = 1N or 11N, D = 5 and so N=0. 2(O55) = 1V10, O can be 5,6,7,8, or 9, and V will be (2O-10)+1 = 2O-9. If numbers can't be repeated, then O isn't 5, it's also not 9 since then V=9, and it's not 7 since that makes V=5. Thus, O = 6 or 8, and V = 3 or 7, respectively.
ODD+ODD=EVEN
655+655=1310
855+855=1710
watching this as a tornado is Goin over my house
Live in NC?
ohio
Ohio... Definitely Ohio.
Ohio is a shit show right now
I slept through it man
Brian got scammed on this so bad. He loses a beer either way. Choose your words wisely!!
As an algebra lover, this was painful to watch.
One does not simply try out the numbers, it's much simpler and more professional to just write down equations (e.g. 2D=N and 2D=E or 2D=N+10 and 2D+1=E) etc.. Eventually solving em all.
By the way for anyone interested there's an app called cryptogram that gives a few of these problems, I think they're really fun.
SPOILER
My best guesses are
655+655 = 1310
855+855 = 1710
I was getting mad becasue i thought there was only one answer and I couldn't logically figure out why both set of numbers couldn't work...then he said there were 2 answers lol
"The V cannot be the same as the D." Remind Bud Light drinkers this everydaym
I'm glad it didn't turn out to be "The square root of 28491 times 2/59 to the power of 8 PLUS ...... = ......"
For 4 minutes I couldn't decide why I kept getting 2 answers
hey Brian! is there any chance you could do a compilation of tricks with pens? I thought about this because I'm in high school and once in class I was thinking "wouldn't it be cool if I were just vanishing a pen repeatedly in the back of the class until someone just turned around and said :how the hell are you doing that?!"
I love how this was uploaded 10 minutes ago and already has about 450 views
Paused the video before the two minute mark and in about a minute got the values. If you haven't figured it out, don't keep reading, but O = 6, D = 5, E = 1, V = 3, and N = 0. Educated guesses FTW!
You can also solve it graphically, although its a bit mathy...
First do some algebra and rewrite the problem:
2(ODD)=EVEN
2(100*o + 10d+d)=1000e+100v+10e+n
200o+22d=1010e+100v+n
Next we know that the one's place of d+d has to equal n because it is the last letter, we can write this algebraically as follows:
mod(2d,10)=n
We also know that e has to be 1 based off of the logic explained in the video
So plugging these into the above equation we get:
200o+22d+1010 + 100v + mod(2d,10)
We now have three variables but we know that o can only be 5,6,7,8,9
and so we can write an equation for each of these values
Now we have 5 equations which we can graph as a function of v
For simplicity I have made the following substitutions:
v=y, d=x
This gives us:
y = (200*o +22d - 1,010)/100
Graphing this, as I did here: www.desmos.com/calculator/dreddhgqse
allows us to visually see that the only integer solutions to the equation required x to be 5
when x (which is d) is 5 in each of our six equations we see that y (which is v) must be 1,3,5,7,9 when o is 5,6,7,8,9 respectively
We now know that these are the only possible answers and they give us the following
555 + 555 = 1110
655 + 655 = 1310
755 + 755 = 1510
855 + 855 = 1710
955 + 955 = 1910
Finally, we can sort through these to see that they satisfy o=/d=/e=/v=/n
and we are left with
655 + 655 = 1310
855 + 855 = 1710
And, well, this is exactly what they found in the video... but at least we have a proof for it now :p
Love the vids Bri Guy!
I had it before they started working it out but then you said to focus on odd and even so i started thinking it was wrong because 6 and 8 weren't odd...i still got it in about a minute and a half so I'm happy anyway xD
Figured it out in about a minute...
I solved this really fast. The numbers I used were 0, 6, 5, 1, and 3. They are in no particular order, but those are one solution.
I solved this in less than a minute... and I kept yelling THIS IS IMPOSSIBLE!!!! because there wasn't just one correct answer. Gah.
O=6, D=2, E=4, V=3, N=1 was my first thought, then they went on about each letter representing a part of the number instead of multiplication
Awesome to see Fraser Cain!
Nobody saw the letters on the refrigerator door that said "hope yall don't degress Russian finance"
Hi and welcome to Baldwatch!
I got to where they got with 655 or 855. I didn't realize having two answers was acceptable so I finished watching. Now I am slightly disappointed lol
Almost got it. In the end I did not realize 955+955=1910 is not a valid answer.
It's an evenly odd number.
If we force rule that different letters correspond to different digits, then we are left only with:
655 + 655 = 1310, 855 + 855 = 1710
It's much easier to solve if you write it like it's long addition:
_ODD
_ODD
EVEN
you instantly get E = 1, so it is
_ODD
_ODD
1V1N
and here we have not much left for letter D: two of them have to sum up to odd number 1, so there has to be carry out from previous position, and it was D as well, so D is at least 5. Also we have to have exactly 1 so without carry it has to be 0, so D is either 0 or 5, but previously we deduced that it is at least 5 therefore it's 5, so now we have:
_o55
_o55
1V1N
after that we get that N = 0 and for o we've got only 4 different values we need to check, 6, 7, 8 and 9, from which only 6 and 8 satisfy condition that all digits have to be different.
Lol, this is pretty much majority of things which children do at 4th-5th grade in my country and this is quiet useful for developing logic and ability to deduce things, I don't understand why it took adults so much time, it's very easy.
Dafuq? They worked it out and Brian talked them out of it.
It took them way too long to figure out the answer.
this was so easy it makes me sad
1:44
Oh, cool, like Sideways Arithmetic from Wayside School! (In fact, they probably used this one, but I've since forgotten it.) Okay, let's figure this out...
Two digits added together will never be greater than 18, so if you carry anything, it will be a 1.
..ODD
+ODD
_____
EVEN
With that said, there was nothing in the thousands column, but we ended up with E. So we know E is 1:
..ODD
+ODD
_____
1V1N
Now, looking at the tens column, we know D+D should be even, but there's a 1 under that sum. So there must be a 1 carried over from the ones column, and D+D ends in 0.
But the units column, which we just said must carry a 1, is D+D. Since D+D both ends in 0 and carries a 1, it must be 5+5=10 (which also means that N must be 0).
..O55
+O55
_____
1V10
Now we know there's a 1 carried over to the 100s column, (and that the result of the sum "carries" over to the 1000s column), so O+O+1=V+10. Rearranging a little, 2*O=V+9. Now if V were even, V+9 would be odd, but it's obvious that 2*O is even, so V must be odd. (And so 2*O=V+9 an ODD no. plus an ODD no. equalling an EVEN no., reminiscent of the problem itself!)
I assume the digits are different (other comments seem to confirm that he eventually says this), so V can't be 1 or 5. That leaves 3, 7, and 9. We'll have to test these through brute force.
Solving for O, we know O=(V+9)/2. That should give us a good starting point. And the key to eliminating possibilities is that all letters are different digits, so if one letter (in particular, O) is the same as another, we have a contradiction and our premise must be wrong.
If V=3, O=(3+9)/2=6
If V=7, O=(7+9)/2=8. That could be.
If V=9, O=(9+9)/2=9. But then O would be the same as V, so that can't be.
So I can't work it out right now, but I've only started the video. Maybe there's a rule that none of the letters are in the names of the digits they represent? Then it would have to be
..855
+855
_____
1710
but I haven't heard all the rules yet
If they were allowed to do doubles, which Brian said wouldn't matter (2:40), then 755 + 755 = 1510 and 955 + 955 = 1910 are both legit answers
I just looked at the d and said ok what double digit will set me up like this at the end. idk I'm good at math but I actually did 55+55 first and I figured out there was actually multiple answers before I even finished the puzzle but I chose 655.
this was so easy. got it within 2 minutes.
1. E must be 1 because two 3-digit numbers added together to create a 4-digit number must be at least 1000, but _cannot_ be 2000.
2. (Since that makes the 3rd digit of EVEN also 1), DD + DD must be 55 + 55 because it is the only combination of two 2-digit same-digit numbers (11, 22, 33, 44, 55, 66, 77, 88, 99) added together that results in a 1 in the tens column.
3. Therefore, N must be 0, since 55 + 55 = 110.
4. O + O + 1 (carried from DD + DD) is at least 10 (as EVEN is 4 digits), so O must be at least 5. There is not enough information to solve, but possible solutions that do not repeat any digits from other solutions are 6 and 8. (655 + 655 = 1310, 855 + 855 = 1710). 555, 755, and 955 do not work because they repeat digits: 555 + 555 = 1110 (that obviously repeats a few), 755 + 755 = 1510 (D and V would both be 5), and 955 + 955 = 1910 (O and V are both 9).
So for those doing this at home and are reading the comments before watching the full video, two important facts that weren't mentioned initially:
1) Each letter represents a _unique_ integer.
2) There are two solutions.
(If the letters don't represent unique integers, there are four solutions).
he doesn't have to state that each letter is a unique integer as that is very obvious else there would've been hundreds of solutions such as 666+666= 1332, 667+667= 1334 etc
the ads are getting smarter
Brian Brushwood Whats the difference between russian finance and american finance, wouldnt they be the same aside from curancy conversions
Brian, where/when do you announce the winners of the throwing cards giveaway thingy.....
posting before answer is shown but i got it figured out O=Any # > 4, D=5, E=1 N=0 ,and V is a little tricky but first i will create a new variable x which is for solving for v, x = 9 - O, And v = O - X
I wrote up a 28 line program that solved it in 0.11389565467834473 seconds.
Says, "algebra question", then has lets represent digits with implied concatenation?
Seems to me, it should be multiplication if you wanted to claim it is an algebraic statement.
I figured it out in two minutes of the video
The equation for this problem is E times the square root of 3 plus V times the square root of 3 plus N times O = 2 times D plus 2 times D. This means only 855+855=1710 can be the answer. The other, 655 is a false positive.
I got the 655+655=1310 Answer in a random way of logic. I am surprised I got it actually.
Wait. I got odd + odd = even as (o+d+d +. o+d+d = e+v+e+n) and using letters as numbers (a=1, b=2) I got 23 + 23 = 46 which totally works too..
Not really algebra, more like a simple cypher, if it was algebra you would simplify to 2O(D^2)=(E^2)VN
But it's actually 200×O + 22×D = 1010×E + 100×V + N
Problem: Find all combinations of digits that satisfy this pattern, each number must be different:
ODD + ODD = EVEN
O=6,8
D=5,5
E=1,1
V=3,7
N=0,0
Or:
655 + 655 = 1310
855 + 855 = 1710
I believe these are all the solutions.
Logic:
First e must equal 1, because it's two three digit numbers, and there is no way for three two digit numbers to be added to produce a number greater than or equal to 2000. (999+999 = 1998, the highest we can go). Therefore, the target number is bounded between 1000 and 1998.
For O since the answer is bounded between 1000 and 1998, we know O must be greater than or equal to 5, and less than or equal to 9. That's all we know for now, we'll come back.
For D, we need to produce the 1v1n pattern, so the 10s place must have a 1 in it. The selected 0 doesn't affect the 10s place, it only affects the 100s and1000s place (we can convince ourselves by this line of reasoning:
ODD + ODD = 2*ODD = 2*(O*100 + D*10+D) = (200*O + 20*D+2*D), so we can clearly see for the 10s place and 1s place, we see that O will be multiplied by 0. and will have no effect, but D will.
. That means DD*2 must have a 1 in the 10s place. We can try all 10 numbers if we want (00,11,22,33,44,55,66,77,88,99) to convince ourselves that 110 is the only number that will produce this, so D *must* equal 5, since 55*55 = 110.
Okay, now that we've got that sorted, we know N has to be 0.
Then V depends on our selected O. Again, we try all 5 values of O (5,6,7,8,9) and we find that the only ones that have a digit in the V place that isn't already taken are 6 and 8.
Edit: I just saw their solution part of the video, just to explain why O=9, O=7, and O=5 don't work (for the sake of completeness of my own solution), it is because the digits are already used. D is already 5, so O can't be that, O=7 doesn't work, because it gives 1510, but then V=D=5, which isn't allowed, and O=9 doesn't work because it gives 1910, so then O=V=9, which isn't allowed because they say you must have a different number of each letter. :)
hacking the system bring it back
For those interested in a mathematical proof of the solutions, I came up with this (feel free to correct me):
Let's find the digits O,D,E,V,N making
ODD + ODD = EVEN
Because the two same 3-digits numbers sum up to a 4-digits number, we know that O ≥ 5. We also know that two 3-digits numbers can't sum up to more than 1998 (in general x-digits number summing up to a (x+1)-digit number will be 1-something). Which means that E = 1.
Let's suppose that D < 5. Then
D + D = N
D + D = E
But that's impossible, because all digits are different. Which means that D ≥ 5. This means that
D + D = 10 + N (N is the last digit of a number between 10 and 19)
E = (N + 1)%10 (that's D+D plus the 10 of the previous one)
EV = O + O + 1
So far we have (solving the equations above)
E = 1
N = 0
D = 5
This leaves us with O and V. We know that O ∈ {6,7,8,9}. If we add 1 to the double of each of these numbers, we get V. So V ∈ {3,5,7,9}. But each digit is unique, which leaves us with
(O, V) ∈ {(6, 3), (8, 7)}
Which leaves us with two only possible solutions
655 + 655 = 1310
855 + 855 = 1710
Please do correct!