Can You Figure Out The Special 6 Digit Number?

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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[MindYourDecisions]

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.


Plus, I bet I'm the only YouTuber 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:


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?

[NestorAbad]

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<=165987:
d1=digits(n)
if len(d1)==6:
if digits(2*n)==d1:
if digits(3*n)==d1:
if digits(4*n)==d1:
if digits(5*n)==d1:
if digits(6*n)==d1:
print(n, 2*n, 3*n, 4*n, 5*n, 6*n)
n=n+1

[jiangling]

Abcdef=
A big chicken defining einsteins formula

[LughSummerson]

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.

[Katpasniss]

I know some special 6 digit number
Like 177013, 228922 and many more

[DarkZeros]

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

[KitsuneZeta]

(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.

[paulobouhid]

*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 . 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 = = 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.*

[Geigenzaehler]

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

[MrDj]

Which one of your books are good for a 53 year old who is about to learn and understand college algebra?

[interruptedtales]

I'm not good at math, but ok with logic.

First, a must be 1 because if a => 2, 6*a >

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.

[jbtechcon]

Interesting, the next row would be 999, 999

[emperor_woooooo]

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.

[omargoodman]

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.

[jagmarz]

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))?

[Skippyrd]

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.

[juregams]

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.

[benhermoni]

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!

[koenth]

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?

[mattzahara]

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.