Can You Figure Out The Special 6 Digit Number?
ฝัง
- เผยแพร่เมื่อ 6 ส.ค. 2024
- A 6 digit number abcdef has the following property. Make a table with 6 rows, where column i = 1, 2, 3, 4, 5, 6 is the number abcdef multiplied by i. (If the first row was 123456, the second row would be 123456 x 2 = 246912). The result is a Latin square: each row and each column has those same 6 different digits. Can you figure out a value of abcdef?
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On my blog, "melted" posted this comment about why the solution is unique, and it proves another way to derive if you didn't know the trick. I haven't checked all the details but it seems correct to me and nicely explained! Here is the comment:
142857 is the only solution.
1 is the only possibility for the first digit--if it were higher, then multiplying by 6 would yield a seven-digit number, not six. (And it can't be 0, because then you need a 0 in the last column somewhere, and the only ways you get that are by making f an even number or 5--neither of which give you 6 different digits in the last column.)
For the last digit, f... it can't be 1, because that's already been used. It can't be any even number, because multiplying an even number by 6 will give you a number with the same digit in the ones place, and we can't repeat. It can't be 5 because then you'll get repeating 5s. So, you only have to check 3, 7, and 9. Of those three, only 7 results in a 1 anywhere in the last column, so 7 is the only possibility for f.
7 being the last digit tells us that the remaining digits are 2, 4, 5, and 8, and it also tells us what order they're in for the rightmost column. So, next you check e, plugging in all four remaining values in turn to see which results in the correct digits in that column. Only 5 does. Similarly for d, only 8 results in the correct set of digits in that column, and for c, only 2 works.
So, 142857 is the only possibility because 1 is the only possibility for a, which makes 7 the only possibility for f, and so on at every turn.
--------
On an unrelated note, TH-cam has partnered with some Creators to encourage digital literacy skills. You know it should have picked me! I am cited in many books, academic papers, and I myself have published 9 books. Plus mathematics is the key to teaching logical thinking and encourages people to find mistakes. Like in this video only careful viewers will find the 3nd mistake. News release: youtube.googleblog.com/2018/07/building-better-news-experience-on.html
Plus, I bet I'm the only TH-camr to point out something the WORLDWIDE media missed, which was the correct answer to the captain's age question which went viral a few months ago:
th-cam.com/video/uyS1cXrsgIg/w-d-xo.html
I've also been covering viral math puzzles so people are no longer confused by fake news spreading the wrong answers. So I ask again: who is better than me for promoting digital literacy skills?
MindYourDecisions its 3rd not 3nd (I guess that's a mistake.)
2:21 3nd
2rd mistake regarding the number 3 after 12323 in the cat-mouse puzzle
presh should make a special video on the number 3 if he blunders with it the 3nd time
If you hear a proof, would you make a video on the proof and a similar problem?
Jess Stuart 156427 has two 2s in the third column,
and 076923 has two 3s in the first column...
Jess Stuart I guess you might've compared only the first row with the others, not comparing them among themselves
Abcdef=
A big chicken defining einsteins formula
To multiply Big Chickens you'll need A Big Cockerel Doing Exciting Fornication. 😉
Quite challenging if you don't know the special property of 1/7. Thank you for sharing!
EDIT: it can be easily tested that this is the only 6-digit number with this property. Here is a little piece of Python code that does it (I'm not experienced in Python at all, so for sure the code may be much more optimised) :
def digits(n):
return set([d for d in str(n)])
n=102345
while n
I give a proof above.
Or, in PERL:
O: foreach my $n (0..999999) {
my @c = ( );
foreach my $m (1..6) {
my @r = ( );
my @d = split //, sprintf("%06d", $n*$m);
foreach my $d (0..5) {
next O if $r[$d[$d]]++ || $c[$d][$d[$d]]++;
}
}
printf "%06d
", $n*$_ foreach 1..6;
}
yes, it is.
I think that it must be "while n
You're totally right, thanks for pointing out the error! I'm editing the code.
2:21 lol 3nd
Pronounced, "thirnd"
Is that better or worse than 3st?
im 3st’y weres my pepsi!
Cédric Wiederkehr maybe
😂
You can work this out without knowing the fraction thing, but you must know what a Latin square is, which Presh did not explain in the video. A Latin square means that the same 6 digits are in each row and column, not any 6 different digits. It's like a number anagram.
So if any line or column is 123456, that means there must be no 7, 8, 9 or 0 anywhere on the grid. And if you work out that there is a 6 in any line or column, then you know there is a 6 somewhere in each line and in each column.
That's enough information to work it out logically. There is only one solution because you arrive at it by elimination of all other options.
I worked it out without knowing either the 1/7 factoid OR the meaning of "Latin square." Not knowing the second just meant there were other numbers to test.
Thanks for the tip, now it's solvable!
R3Testa I'm curious, are there solutions where there are unique digits in every row and column but they aren't the same 6 digits?
chinareds54, The number 142857 is the only solution, even if the set of digits in each row (or column) don't have to be the same six digits.
well the 1 in the first column is a given, so there are 5x4x3x2 combinations left, which actually isn't that bad the brute force way but why bother. Not a fan of this puzzle.
Logically I can figure out this:
a=1
b=2/3/4/5/6 otherwise will overflow
f=3/7/9 otherwise last digit will repeat
DarkZeros Tried the Same way; still No Clue What the fraction of 7 has to do with this 🙈
f can also be 1
Aman Kumar Singh no, it can't, because then there would be a repeat in the e column.
Yep that's what I got. I was doing process of elimination for what the possible first and second digits are. I gave up after ten minutes.
a 6 digit number could include the special case a=0? ;-)
Interesting, the next row would be 999,999
Which, fractionally, is the same as 7/7, since 0.999999... == 1!
Excellent observation!!
> 1!
1! = 1
beep boop i'm not a bot
The interesting stuff about that is that it is not exactly equal to 1, so 1/7 is not exactly equal to 0.142857. indeed, it is repetitive: 0.142857 142857 142857 ...
Hey there, your 6 digit numbers links to this hentai manga:
[Kimura Neito] Kuroki Hitsuji no Dinner Show
And I gotta say you have some fine tastes in hentai. I salute you sir.
sadly it is not translated, the art is really good tho
(this was worked out at 1:01 in the video)
Before delving deep into the solution, there is an observation that can be made: First off, because the number in the sixth row MUST be 6 digits long and it has a value exactly equal to the first row times 6, the number represented by "a" MUST be 1 (as if it were 2, then we'd have 7-digit numbers in the 5th and 6th rows, and if it were 0, it would become impossible to satisfy the Latin Square requirement [this is a trivial proof - set B to 9, and note the five numbers that would appear in Column A]). You can start by writing that number in first, but leave the rest of that column empty.
The remainder of the square can be sussed out by trial-and-error, beginning with column "f". Because column "a" MUST be 1, column "f" MUST be an odd number that, when multiplied by any ONE number from 2 to 6 and modulo divided by 10, produces a result of 1. The only number that satisfies this is 7, which when multiplied by 3 is 21, and when 21 is modulo divided by 10 it produces 1. Fill in column f with the appropriate numbers (the numbers to keep in mind are are 7, 4, 1, 8, 5, and 2)
Because column f's results do not depend on any other column's results (the constraint on needing a 1 is due to knowing column a must be a 1), we now know the six numbers that MUST be in each row and column. We've already used two of the numbers (1 for a, 7 for f), so we can now solve for "e". The method is similar, but with an additional step: we have to take into account the tens digit of the number from the column to the right (e.g. if we were to try e = 2, for the X2 line we would have a result of 2 * 2 = 4, + 1 from the tens digit of 7 * 2 = 14, producing a final value of 5). The number 5 will actually satisfy all conditions when placed in column e.
Applying these steps to columns d, c, and b will produce solutions of d = 8, c = 2, and b = 4 (note that you do need to factor in ALL numbers to the right of the column you're working on, not just the immediate column to the right).
Therefore, the number abcdef is 142857. The final grid is as follows:
142857
285714
428571
571428
714285
857142
Because there is only one possible value for "a", and that value for "a" makes only one value for "f" possible, this is the ONLY solution to the puzzle.
Kitsune Zeta Why Couldn‘t f Be 3 or 9 at the start
Thank you for supplying what was missing from the video. "I just used the fact that the fractions of 7 are cyclical" felt a bit too hand-waving to me.
Once you've found f, you can also fill in the left column since the numbers will be in ascending order, so their most significant digits also have to be ascending. That gives you b (2*ab
We know that 1 is one of the digits. That means that there has to be a 1 in the F column. To get the values for column F, you can take the initial F value and multiply it by 2 3 4 5 then 6 to get each additional F value. If you use 3, you get 6 9 2 5 and 8 in the rest of the F columns. That doesn't have a 1 and we know that one of the Fs has to be 1. If we use 9, we get 8 7 6 5 and 4 in the rest of the F columns. That doesn't have a 1 either. If you use 7, the next F column numbers are 4 1 8 5 2. Notice this time we finally have a 1. That means that the F column must be 7 4 1 8 5 2 from top to bottom.
I have a big problem with your explanation in the beginning.
The only property is, that every row and column has no repeating digits, am I right.
But a roman square is different. It has the additional property of having the same digits in every column/row.
Sure most do know what a roman square is, but your wording makes it sound like you were altering the properties of a roman square. "in other words" creates an equivalence that is not appropriate.
I checked it with code: If you take the loose restrictions without having the same digits in every column/row, there is still only one solution, but others get very close: Take n = 156427
156427
312854
469281
625708
782135
938562
The only double digit is the two in the 3rd column.
I am sure there is no quick proof, that n = 142857 is the only solution.
Here are all numbers, where only one digit is doubled:
126857
253714
380571
507428
634285
761142
126873
253746
380619
507492
634365
761238
127309
254618
381927
509236
636545
763854
130869
261738
392607
523476
654345
785214
142873
285746
428619
571492
714365
857238
154273
308546
462819
617092
771365
925638
156427
312854
469281
625708
782135
938562
*Solution to those who didn´t know about the cyclic property of 142857:*
Let´s call the number *N* .
Since the digits are all differents and there is no "0", its digits will add up to a number between *21* (1+2+3+4+5+6) to *39* (4+5+6+7+8+9), which we will call *S* (a two-digit number).
If we add the digits *in each column* of the 6x6 matrix, we get *S* as the sum, and so adding the 6 numbers we will have *SSSSSS* . But the sum of all the 6 numbers is N + 2*N + 3*N + 4*N + 5*N + 6*N, which equals *21*N.*
Thus: 21*N = SSSSSS = 111,111*S, and N = 5,291*S.
Multiplying each *S* from 21 to 39 by 5,291 (in fact we need to test only the *odd* ones, since N must be odd), we will get, as result, all the numbers containing repeated digits or a "0" (even both), the exception being S = 27, and we will get the unique answer *N = 5,291 * 27 = 142,857.*
Which one of your books are good for a 53 year old who is about to learn and understand college algebra?
Is there an easy way to understand it better?
I'm thinking maybe a College Algebra book?
DERRICK JAMES - I highly recommend "Paul's Online Math Notes." They introduce the principle, provide an example, give a clear step-by-step solution, then allow you to practice until you've mastered it.
"Paul's Online Math Notes" has every level of math: algebra, trigonometry, calculus, differential equations. It's a free online resource that got me through college calculus.
Hey I solved it just like you Presh! Knowing fraction properties for the win
One in seven knows the answer immediately.
By the explanation, if someone didn't happen to know what a Latin Square was (I didn't know it by name but I knew the principle of a fixed set of _n_ digits arranged in rows and columns without repeats), they would presume that you were establishing the definition of a Latin Square as just having no digits repeating in any row or column but not necessarily that it is the same fixed set of digits arranged in each. In other words, your explanation didn't make it clear that the combination of digits had to be consistent among each row and column. This made it so it wasn't a test of logical progression but just a matter of mathematical trivia; if you happened to know, off the top of your head, the cyclical nature of 7's, that gives you a valid solution. It would have been a lot better had you explicitly defined the term _Latin Square_ in the intro so everyone would be on the same page about the parameters of the problem.
I was so confused on why I needed to find abcdef. Was each row supposed to have different values?
I think the explanation could have been clearer. It wasn't immediately obvious that you're multiplying abcdef for each row. My first analysis was that you're taking the previous row time the new number, which follows the example given, but of course can never work since you end up with zeroes under digit 'f' for at least the last two rows. Would have been better if the example had been for some row other than 2.
From logical analysis, 'a' must be either 1 or 0, and 'f' cannot be even, or 5. That's as far as I got before I bailed and wrote some quick code to solve it. I didn't know the fractions of 7, but I did see that all the rows were the same numbers, in the same order only shifted. The explanation for this phenomenon is really neat! Does anyone know whether all solutions to problems like this end up being multiples of the same fraction (1/(n+1))?
Yeah ok, but if you don't know the fact about fractions of 7. How to solve this?
I noticed immediately that 'a' has to be 1, or another tens column will need to be added. 'F' can only be 7, as all even numbers can be ruled out as candidates (not resulting in '1') and the other single digit odd numbers don't have multiples ending in '1' until multiplied by a number larger than 6 (3: 3,6,9,2,5,8,1,4,7),(5: 5,0 etc),(9: 9,8,7,6,5,4,3,2,1). From there the 100,000 column can be ordered from smallest to largest: 1,2,4,5,7,8. The 10,000 column can now be easily filled in, 'b' cannot be 5,7 or 8 as multiplying that by 3 would result in carrying the 1 to the 100,000 column. 2 can also be ruled out as multiplying that by 4 does not result in carrying the 1, which it has to. Therefore 'b' is 4. Everything after that kind of fell into place.
I agree with your disappointment. I wrote a proper reasoning-out of the solution above. It also proves the uniqueness of the solution.
theonesparrowhawk Maybe the definition of a Latin Square states that all rows and columns need to only contain abcdef, but as he described the problem that restriction doesn't exist. Just that there are no repeats in a row or column, in which case I don't think that you can easily discount f=3 or 9
Latin Square: 'an arrangement of letters or symbols that each occur n times, in a square array of n^2 compartments so that no letter appears twice in the same row or column'. In this case n=6, and as such both 3 and 9 can be ruled out using my reasoning above.
theonesparrowhawk again, the problem doesn't state that; you have to look up the definition of a Latin square or know it beforehand to know.
I'm not good at math, but ok with logic.
First, a must be 1 because if a => 2, 6*a > 999999
Second, f cannot be 1. Further it must be odd because every even number will repeat their initial digital over 6 multiplications. 0*6 = 0, 2*6 = 12, 4*6=24, 6*6=36, 8*6=48. Trying odds that are not 1, 5 is out (5*3 = 15). We therefore have 3, 7 and 9 as possibilities. 3/6/9/12/15/18, 7/14/21/28/35/42, 9/18/27/36/45/54 . Further, we know of these that f must be 7 because one of the final digits of f*1..6 must be a 1 in order to repeat the 1 in the final f column, and only the multiples of 7 fit this.
From there we can place the multipled value of f in each column.
a b c d e f
1 x x x x 7
x x x x x 4
x x x x x 1
x x x x x 8
x x x x x 5
x x x x x 2
And then futz about a bit, which I believe is the technical mathematical term, to swap the remaining four digits (4/8/5/2) to make sense and not give repeats.
Not every row, just the "f" final row (the last digit). The other rows can have "carry"s added to them. The "f" row can't because there's no digit to its right.
Sorry - I mean to say the other *columns* can have carries on them, while the f column can't.
He never said that the other 5 rows had to use the *same* six digits as the first row, so there could be a 3, 6, or 9 in the last column and there is not necessarily a 1.I think this means you have to try all three candidates for f, so triple the amount of futzing.
But then he goes on to state that this is a "Latin square" which implies that only 6 distinct digits can be used throughout. If you had to know what a Latin square was to understand the statement of the problem, the long song and dance about the numbers being unique in the rows and columns could have been skipped.
Note also, though, that b can't be larger than 6, and if b is 6, c must be 5 or less (again because 167xxx times 6 > 999999), which reduces the futzing a bit.
I'll point out an error and an inefficiency in your proof. First, at the beginning a could have equalled 0, which would also avoid overflow; but yeah, your next step would rule out a=0 anyway, so no harm done. But once you got f=7 and filled in the f column, you could ALSO have deduced that the a column had to be the same digits as in f but IN ORDER. :-)
How can i send a problem for your channel? pls help
Can you show a solution that doesn’t pluck a result out of thin air? I was trying to logic through this, and then realised there would be a lot of cases and I couldn’t be bothered.
I like the quicker intro!! I think quickly tho. I like the speed of the question as well, but might be a bit fast for others. I really like the newish countdown as well, buy maybe instead of swishing the count down 3-2-1 you say something about your books during that time.
If you take the slightly expanded problem where the rows must be the last 6 digits of 2x, 3x, 4x... of abcdef, there is exactly one other solution. 854273
708546 (trimmed leading 1)
562819 (trimmed leading 2)
417092 (trimmed leading 3)
271365 (trimmed leading 4)
125368 (trimmed leading 5)
Not sure if this solution is particularly interesting, solved for it computationally.
To solve this question I have created a program to check every possible number (in a way that the numbers are 6 digits and if you multiply them by 6 they are less than 7 digits.) It took me about 30 minutes to program it, just to figure out the one possible number - 142857, only to see you could solve it instantly if you knew that the fractions of 7 are cyclical. Great question though!
I suppose this is redundant but surely there are different solutions in other bases, the puzzle does not specify a number base.
I figured it out, but used brute force and recursion. Didn't take too terribly long, as the numbers get eliminated pretty quick if you start from f and work up.
That was great! I could only think that "a" must be "1", because any other number would be higher than "9" when going through the multiplications, and this situation leads to numbers with more than 6 digits.
Write the number 142857 on a short piece of ribbon, then tape the two ends of the ribbon together. To multiply the number 142857 by 2, 3, 4, 5, or 6, you can cut the ribbon between the appropriate pair of digits.
I know some special 6 digit number
Like 177013, 228922 and many more
[Kimura Neito] Kuroki Hitsuji no Dinner Show
LMAO
It was unclear that the same 6 digits would be on all the rows from the intro.
By intro, they don't have to be the same. This just makes the problem a bit harder to solve.
Stop making excuses.
If you had been given that information is would have simplified the challenge.
It eliminates much more of the search space. You know 1 is the first digit (anything else exceeds 6 digits with 6x). You know the final digit must be odd (multiples of even numbers are even, so you only have 5 possible final digits but need 6). Among the odd numbers, you can eliminate 1 as the final digit (already in space 1) and you can eliminate 5 as the final digit (repeats at 3x and 5x). If you don't know the definition of a Latin square requires the same 6 digits have to be used, you have a search space that includes 1bcde3, 1bcde7, and 1bcde9. If you know the Latin square re-uses the same 6 digits in each row, you can eliminate 3 and 9 because 1 doesn't appear in those until you get 3x7 and 9x9, which are both outside the square.
Knowing the additional feature of a Latin square reduces your search space to ~1/3 of what it is without that tidbit of knowledge.
+Michael Onines
The compression is much more than that, because even the 1bcde7 case will branch further without the additional constraint.
Quick Note:
For those that don't know what is a _Latin square_ you should have mentioned in the video, the same you wrote in the description: "each row and each column has those *SAME* 6 different digits"
This is fun puzzle. I figured a=1 instantly. Since 11, 31, 41 are primes and 51/6 isn't intiger the only way to get 1 in the end is 21=3*7, so f=7. Then you figure it out that the remaining numbers are 2, 4, 5 and 8. Then I checked place e for every row and by eliminating I figured it out that e=5 and then by the same method that d=8, c=2 and b=4.
Keep up the good work Presh... The Haters are going to hate... Most uneducated people always hate what they don't understand.
I knew this one instantly as well, thanks indeed to knowing the fractions of 7. Also, I recall this special number being mentioned on the old PBS show Square One TV, where 142857 is special in that it retains the same digits, just in a different order, when multiplied by 2, 3, 4, 5, or 6. (And interestingly, I noted that 142857 x 7 = 999999.)
Well, I didn't know the cyclical property of the 1/7 fraction so I took a while (something near 20-30 minutes) to solve it. I worked around roughly as follows: I figured out that a=1, cause if it's any digit other than 1 then the number either becomes a 7 digit number when you multiply it by 5 or 6 (a >= 2) or if it's 0 the column would have at least another 0, which isn't possible by the Latin Square property. Then I started thinking about the last digit f: now, since I know the property of the Latin Square that's hidden in this description, which is that any other digit in the others rows/columns can't be any digit but has to be one among the {a,b,c,d,e,f} digit set, I knew that the column in which the f digit appears has to have exactly one "1" digit (and exactly one "f" digit of couse). I tried few numbers and I realized the only one who makes the "1" appear in the column just once if you multiply it by 2, 3, 4, 5, 6 is the "7" digit, so f = 7. This gave me another precious information: since 7 has to be the first digit, I figured out the other possible digits just by making the products of 7 with the 2, 3, 4, 5, 6 factors, and I found out the possible set of digits is S = {7, 4, 1, 8, 5, 2}. From there I just started to make the "partial" products with the digits I know with each factor listed to find out the "equations" that could identify the other digits. For example, I know that e has to be multiplied by 2, then 3, then 4, then 5 and then 6 and I know I have to add to that result the "report" of the product of 7 too, so I found out 2e+1 (product with 2), then 3e+2 (product with 3), 4e+2 (product with 4), 5e+3 (product with 5), 6e+4 (product with 6) and I started to try the numbers in the set S and that's basically the process iterated for each subsequent digit. It's a bit complicated but surely doable. Very nice puzzle. I'm sorry if my explanation isn't clear, english is not my main language.
"This gave me another precious information: since 7 has to be the first
digit, I figured out the other possible digits just by making the
products of 7 with the 2, 3, 4, 5, 6 factors, and I found out the
possible set of digits is S = {7, 4, 1, 8, 5, 2}."
Hey, that's neat! I was long aware of properties of 1/7, and I also considered the multiples of 7 mod 10 during solving this puzzle (and many other puzzles), but I've never realized before that the first six multiples of 7, mod 10, correspond to the repeating decimals of 1/7 .
Awesome solution
I have one issue with the problem as initially stated: It does not specify whether the six digits are the same in each row.
I got lazy...Used a script to search the space. You can say that abcdef>50000 because otherwise abcdef*2 < 100 000 which would make first digit 0 twice. You can also say abcdef < 166666 because otherwise abcdef * 6 > 1 000 000. Search space of less than 117k numbers. Additionally, you can immediately toss out any candidates for abcdef in which the digits are not unique making the search space down to (2*9!)/4! which is approximately 30k numbers. Small enough search space for an algorithm to find the solution in a timely manner. Yes its brute force but sometimes you want a fun programming exercise to keep your skills sharp :)
Took me 5 seconds looking thumbnail to understand that we were talking about the property of 1/7, love that number
I totally didnt think about the 1/7 property! But i know that a can only be 0 or 1 since 2×5 or 2×6 will go to 7 digits. Then, checking for the one's place digit for x × y = ending with 0 or 1, den finding y and all the other last digits possible and there u go, u found it.
a solution through induction, rather than rote knowledge:
we start with an upper bounds:
x*6 < 100000 (to stay a 6 digit number in the 6th row)
x < 166666, and because digits can't repeat x < 165987
this means that A must be either 1 or 0
now we turn to F:
The 1s column for F*N from 1 to 6 must be unique, which removes all even numbers Y*6 = ?Y for all even numbers, and 5 is out because 5*N always ends in 0 or 5.
now we are left with 1, 3, 7, and 9
the sequences for 3 and 9 do not contain 1 or 0, so we can remove those, as we already know that A is either 0 or 1.
the 1 sequence is out because if F=1 than A must be 0, and 0 is not in the 1 sequence.
Via process of elimination, we have found that F=7 and A=1.
Now we have the correct sequence (7, 4, 1, 8, 5 , 2,) and we've solved for A (1) and F (7), we can solve for E-B sequentially. We know that E is in the group (4, 8, 5, 2) and that after adding the carries from F (0,1,2,2,3,4) we can figure out which of the 4 remaining values should be E
E=2? -- 2, 3, STOP. E cannot be 2 as 3 is not in the sequence
E=4? -- 4, 9, STOP. E cannot be 4 as 9 is not in the sequence
E=5? -- 5, 1, 7, 2, 8, 4, DONE. E could be 5
E=8? -- 8, 7, 6, STOP. E cannot be 8 as 6 is not in the sequence
repeat this process for D, C, and B, and you'll reach the solution pretty quickly.
"the sequences for 3 and 9 do not contain 1 or 0, so we can remove those, as we already know that A is either 0 or 1."
This step can only be made when assuming that each row contains the same six digits. If one doesn't make that assumption (and also ignores Presh's remark about Latin Squares), the problem becomes much harder; but 142857 will still be the _only_ solution.
I'm gonna take a wild guess and say it has to with the repeating digits of fractions with 7 as the denominator, cuz it's a sequence of 6 unique digits that just keeps getting shifted around every time u multiply. So 142857?
"Can you figure it out?"
*..No*
Perhaps give us challanges that we can logicly work out without knowing these kinds of things suah as knowing the properties of 1/7
These kinds of questions appear at mathematics competitions where either you know the trick or theorem, or you don't. Granted, some problems can be solved in multiple ways with the "trick" making the problem nearly trivial.
I agree with your disappointment. I wrote a proper reasoning-out of the solution above. It also proves the uniqueness of the solution.
If you look at the far right column (1s digit) you can convince yourself that the value of f cannot be 1, 2, 3,4, 5 or 6. When you get to 7 and then when you look at the multiples of 7 you realize those are part of the 1/7 continued fraction. It's a trick, sure, but a trick we learned pretty young, eh?
I was scratching my head over how I rejected a value of 3 for the f digit. Turns out, ab has to be less than 17. So a has to be 1. 3 doesn't produce a multiple that has a 1 in the 1s digit until 3x7 which wouldn't be in this square - so f cannot equal 3.
This absolutely can be worked out without the 1/7 trick. I did so, starting with a having to be 1 (2x5=10), b having to be 6 or less (7x6+60=102), and f having to be 3, 7, or 9 (to not repeat itself).
Note that 3 cannot be eliminated by simply knowing there will be no 1 in the f column, as there is no rule stating that the same 6 digits must be used for each row, only that there must not be any repeated digits in any row or column.
I did a python program to check (checked by brute force in 0.75s) and the only solution is the one you stated.
I also tried ignoring numbers with overflow and the same result (that is every row N as (row(N) x N) mod 1000000). If we instead add abcdef to the previous row and strip any overflow (row N as (row (N-1) + row 1) mod 1000000) we also get 854273 as a solution.
Just a quick bit of logic to narrow the possibilities.
The absolute largest number this 6x6 Latin square may contain is 987,654, which would only be valid on the last row.
Knowing this, we may narrow the upper limit of 'abcdef' to 987,654 / 6 = 164,609. This is not a valid option because of the two sixes. Therefor, the absolute largest 'abcdef' may be is 164,598.
The obvious lower limit is 102,345. This leaves a range of 62,254 possibilities. There are obviously less than that after you impose the distinct digit restriction on every number in that series, but you get where I'm going with it.
Honestly, I would've just written a little recursive Python script to solve it. It would be pretty similar to a Sudoku solver if you've ever done one of those.
Great videos Presh! Keep them coming.
Someone's getting real tedious about his name lately
RDVMusic - He said it in the beginning like he wasn't sure. Presh Talwalker?
His name is one of the best things about this channel lol pressure cooker ;)
I figured there was an easy way that I would never figure out so I wrote a computer program. I also figured the easy solution might only be one number and I wanted to see if there was more than one solution. The solution given is the only one.
Quite impressive, I must say. An excellent ability of thinking out of the box. I mean how would a seemingly impossible puzzle be solved just with the help of 1/7!!! Thank you for such great minds of learning.
But yeah it could be solved by using programming algorithm. Although I have not checked for the answer but yeah use of brain is more wiser than use of technology! Sometimes!!
What is the point of solving these numbers via Latin Square Method
Really driving home that pronunciation and spelling these last few videos, Fresh Skywalker.
Lol i like how he has to make his name bigger in the beginning of the video because people keep mistaking it XD.
Preshtal, can you remove the annoying 80's era coin sound effects when your plugs pop up on the screen?
123456
612345
561234
456123
345612
234561
Why get so complicated by knowing the fraction of 7, just convert them into the no.of columns or rows, in this case 6 and shift them a row further to get the answer, It just looks neater and doesn't need any calculation.
Even without the requirement that each column has a-f this is the only solution without repeats in columns. I found it using excel.
The first digit must be 1 otherwise multiplication gives a 7 digit number so I made a list of every number from 100000 to 199999. Next I had excel find only numbers without repeating digits and deleted the others.
I repeated this process with each product which eventually narrowed the list down to 15 numbers. I then checked each column. Some are close (for instance 157427) but 142857 was the only number without repeats in columns.
0:00 Hey... this is Presh Talwalker? You said that like you weren't sure.
Ok so I didn't know about 1/7 but i figured it out (... less instantly though, took me about 10 minutes)
First: a must be either 1 or 0 since abcdef * 6 is 6 digits long.
Second: f can't be even because 6*f would then end with f.
Third: f can't be 1,3,5 or 9 because the multiples of 1 don't contain 0 (and a is either 1 or 0), the multiples of 3 don't contain 1 as an ending digit (within the 6* frame), same for 9, and 5 is repeating too much. Which leads to the last possibility: if abcdef exists, f must be 7.
Fourth: we can deduce from Third that a=1 and abcdef is 1bcde7. The remaining digits to place are: 2458.
Fifth: b must be smaller than 5 since 3 isn't one of the digits of the square, so b is either 2 or 4.
Sixth: if b is 2, c must be 5 or bigger, because 7*2 already ends with 4. So in that case, c is either 5 or 8. However, this leads to an impasse, since then d must be the remaining number bigger or equal to 5, which leads to e being 4, and 4*2+1 is 9 which isn't one of the allowed digits. So b can't be 2 and b must be 4. We now have abcdef = 14cde7
Seventh:c must be smaller than 5, otherwise b*2+1 would be equal to 9. So c is 2 and abcdef is 142de7.
Eighth: We now have two possible numbers: 142587 and 142857. If we multiply the first one by 2 we get 285174, which verifies the conditions of the second lane of the square, but this is not the case for the third lane. So abcdef can't be 142587
Ninth: There is only 142587 left AND it verifies the conditions for all the lanes of the square.
I actually began this trying to disprove the existence of such a number... intuition can be a foolish guide, but methodic work leads to the truth!... sometimes!
This also demonstrates that this is the only solution (in base ten) since it is found by eliminating all other possibilities.
142857
Saying it before the answer comes up.
I find mathematics very interesting and wanted to try to solve this but somehow I was distracted by the way that the image looked like an electric guitar fretboard.😮
I'm a math guy but I honestly don't even want to try this.
Project Euler question 52 prepared me for this :D
Happened to know of the 1/7th's, so I got that going for me, which is nice. Fun property of the sevenths: 1/7=.142857, or 14-28-57 as he says. 14x2=28 and 28x2≈=57 (and 57+57=114, thus the cycle). So 2/7 doubles 14-28-57 into 28-57-14 (.285714), double again to 4/7 for 57-14-28 (.571428). There's also a fun property that I don't know why it works: 14+28 = 42, 28+57=85, and 57+14=71 (42, 85, and 71 each appearing the the 1/7 sequence as the offset digits to 14-28-57), so 3/7 = 1/7 + 2/7, which is 42-85-71 from the previous sums, or .428571. (You can do 5/7 and 6/7 similarly.)
Another fun thing is if you add those first digits and those last digits the sum will be 999, and 7 × 142857 = 999999.
123456/999999 and 1/21 are both 6 digit repeating decimals.
In fact, there are a ton of them: (6 digit # not divisible by 7)/999999
As for proving that cyclical property...
0:35 'In other words, this is a Latin square'. But a Latin square requires in addition that only 6 different digits are used throughout the square.
By which logical argument does this follow from the previous statement 0:28?
agree
Lol this solution reminds me a tutorial how to draw an owl
Where did 1/7 come from
Math Weeb: Put Glasses On
A similar problem, but you multiply the number by 1, 3, 4, 9, 10, and 12 and get one Latin square, and you multiply the same number by 2, 5, 6, 7, 8, and 11 and get another Latin square.
076293 and 153846?
I thought a Latin Square of order n only uses the numbers 1 to n, not nonconsecutive numbers.
I also knew about the 1/7 property.
If the first row was 123456, the second row would be 123456 x 2 = 246912. Then how do you get 428571 in row three. Did you multiply the first row or the second row by three? It doesn't make sense.
Going at it with what is likely the long method:
a must be 1, as any other digit will exceed the bounds.
b6 + carried value cannot exceed 39, so b = 2, 3, 4, 5, or 6.
f must be 3, 7, or 9, as all others repeat.
if f = 3, e != 2.
if e = 4, d !< 5.
if d = 5, c = 8.
b has no valid value, so... d has no valid value
e != 5.
if e = 6, d = 8.
if d = 8, c = null.
if e = 7, and d = 2, and c = 4, and b = null.
e = null. f != 3.
Repeating for f = 7...
1 4 2 8 5 7
2 8 5 7 1 4
4 2 8 5 7 1
5 7 1 4 2 8
7 1 4 2 8 5
8 5 7 1 4 2
And the answer. 142857
I didn't exhaust all other possibilities beyond "if f = 7, e !< 5. With e = 5, d must only be 8."
Nice that you know that thing about the fractions, but you didn't explain how to solve the puzzle if one doesn't know that. You didn't even follow a logical path: you assumed that what you knew was a possible solution and you simply tested it out.
The way to solve it from scratch starts by realising that 'abcdef' must be between 123456 (the smallest 6 different digit number, not counting 0) and 164598 (the largest 6 different digit number which multiplied by 6 is still a 6 digit number). Therefore a=1.
After that you think about 'f'. It cannot be an even digit, or the last number (6xabcdef) will end on the same digit as the first; it cannot be 5, since it makes the last digit in subsequent numbers be either 5 or 0; that leaves 3, 7 and 9. With either 3 or 9, none of the last digits of all 6 numbers will be 1; since we already stated that a=1, then f=7 and the digits must be 1, 2, 4, 5, 7 and 8. We have now:
1 b c d e 7
2 _ _ _ _ 4
4 _ _ _ _ 1
5 _ _ _ _ 8
7 _ _ _ _ 5
8 _ _ _ _ 2
The digit 'b' cannot be 2 (the 3rd number would start with a 3) nor 5 or higher (the second number would start with a 3), therefore it must be a 4. Subsequent digits can be added now, knowing the possibilities:
1 4 c d e 7
2 8 _ _ _ 4
4 2 _ _ _ 1
5 7 _ _ _ 8
7 1 _ _ _ 5
8 5 _ _ _ 2
The digit 'c' must be 2 for the leap of 5 in the 2nd column between the 3rd and 4th numbers. Little by litte we can now fill the whole table, observing the rules of the Latin square and min-maxing the sums.
And with this method you also prove that this is the only solution.
# Python 3.7
# only one solution was found after examining every permutation
from itertools import permutations
digits = list(range(10))
for row1_list in permutations(digits, 6):
get_next_permutation = False
# skip if leading zero
if row1_list[0] == 0:
continue
square = [row1_list] # initialize square with the first row
# convert the first row's list into an int
row1 = int(''.join(map(str,row1_list)))
# loop over the other rows
for i in range(2, 7):
row_n = i * row1
# skip if any of the rows are longer than 6 digits
if row_n > 999999:
break
# skip if the other rows' don't have 6 different digits
row_n_list = [int(d) for d in str(row_n)]
if len(row_n_list) != len(set(row_n_list)):
break
# the row is good so add it to the square
square.append(row_n_list)
if i == 6: # then we have all good rows so test the columns
# skip if the columns don't have 6 different digits
for j in range(6):
col_j_list = [row[j] for row in square]
if len(col_j_list) != len(set(col_j_list)):
get_next_permutation = True
break
if get_next_permutation:
break
# we have a solution so print the square
print("The Latin Square is:")
print(row1)
for j in range(2,7):
print(j * row1)
print()
You really need to put this IN THE VIDEO, we shouldn't have to read the description to get this hint... "The result is a Latin square: each row and each column has THOSE SAME 6 different digits." Your first example of 123456 making the second row 246921 should have explained that was invalid for two reasons: It has a duplicate 2 in the same row, and it doesn't use 3 or 5 that are in the first row.
Thanks i did it
PROOF that 142857 is the ONLY Latin Square Solution: (latin squares have the same 6 numbers in each row and column)
1. A=1. A can only be 1, as any other number will produce double digits when multiplied by 6
2. F=7. The digit 1 must appear in the final column. 7 and 1 are the only numbers which can be multiplied by a number between 1 and 6 and result in a 1 in the final column. It can not be 1 because it already appears in the x1 row.
3. Given that F=7, the F column must be 7, 4, 1, 8, 5, 2. Given that this is a latin square, we now have all the number in the grid, and just need to place them.
4. Column A must be 1, 2, 4, 5, 7, 8 in order as the multiplier increases.
5. B=4. Because the x5 row is greater than 700,000, abcdef must be greater than 140000. Because the x6 row is less than 900000, abcdef must be less than 150000. Therefore, b must be 4.
6. C=2. Because b is 4, the second number in row x2 must be either 8 or 9, depending on what is carried from column c being multiplied by 2. But 9 is not in this latin square. Therefore, nothing is carried from column C. Thus, C must be less than 5. The x1 row must contain all 6 numbers for the latin square, and a=1, b=4, and f=7. Since C must be less than 5, it can not be 5 or 8, and must be 2.
7. E=5. Only 5 and 8 remain for the x1 row. The E column must contain 1,2,4,5,7 and 8. Because F=7, E can not be 8, as 87x3 would put a 6 in column E. Therefore E=5.
8. ABCDEF=142857. D must be 8, the only number remaining in the latin square.
What kind of math meanies are out there?
Vihart?
Aleph Naught :o explain yourself!
The Ruk no explanation is needed
Aleph Naught I know nothing on the situation
I kinda memorized that n/7 ≈ mod(10^(n-1)*1/7, 1) and that 142857/999999 = 1/7 therefore 1,2...6 all over 7 will just rotate through the string 142858, and therefore satisfy what you are asking for.
Typo 142858 supposed to be 142857
Aaaaand that's exactly your solution. Darn, thought this one would be more elaborate
And worse so I had the rule wrong. 10^n was part of some other proof related to why 1/6=0.1666. argh
Hello from France
Oh what a sad person I am, I didn't know what about the 1/7 thing so I used a sledgehammer (AKA Excel)
By inspection I knew the first digit was 1; the second
New intro !!
This video does not really present a way to find the solution - it is based on a special piece of knowledge, i.e., that 142857 is cyclic. A way to figure it out NOT knowing that would be good. I also solved it in seconds, as I knew that number, but have no clue how to tackle it otherwise.
What´s about the negative number as a second solution: -142857
At least it is also a 6 digit number, as well as "special" ;-)
I saw that number in my math textbook so I solved instantly
I knew it! I just knew that 142857 would work!
i knew the fact but still i didnt solve the problem because 1. its not obvious and 2. i didnt understand i thought multiplying the PREV. row by 2, 3, 4, 5, or 6
William Martin Powell That would balloon exponentially beyond six digits quickly, so I immediately dismissed that particular notion.
yup
That's no Parker Square!
sinecurve9999 so if there's a hypothetical solution with all unique digits in rows and columns but NOT the same 6 digits, would that be a Parker Latin Square?
+chinareds54
By the power of brute force, I proved that no such Parker Latin Square exists (unless my code is buggy cough).
Only 142857 is a valid number
Oddity:
on the interval 102347 to 166666, using brute force excel, 11 numbers got 11 of the 12 rows/columns correct, and i mention them:
123089, 126827, 126857, 126873, 127309, 128369, 128427, 130869, 142873, 154273, 156427
Solved this using Python and numpy. Did not know about the cyclical property of 1/7.
The cc says this is pressure walker
When you already knew the answer because it was on Scam School :D
just try every number it wouldn’t even take that long
That's why God made computers.
i was not able to deduce the number. i gave up after a few days.
so the answer is Kuroki Hitsuji no Dinner Show, interesting...
I'm guessing the answer is 142857 ?
Is there any other solution ?
I doubt it.
Wanna know how I guessed ?
Pretty easy,
I remembered that 1/7 is cyclical from the first time i saw this video.
:D.
142857. Before I even watched the video. I saw this puzzle years ago.
I guess this has to do with 1/7, so then abcdef would be 142857
I don't even have to watch. 142,857. My favorite number. (999,999 ÷ 7)