Answer to the Fiendish Second Conway Puzzle, The Ten Divisibilities!

John_H_Conway_2005_(cropped)

Last month, in honor of mathematician and puzzly spirit John Horton Conway, we shared two of his favorite brain teasers and challenged our fellow PuzzleNationers to crack them.

Two weeks ago, we shared the solution to puzzle #1The Miracle Builders, and offered a few hints for puzzle #2, The Ten Divisibilities.

Now that we’ve heard from a few solvers who either conquered or got very close to conquering the second puzzle, we happily share both the solution and how we got there.


The Ten Divisibilities

I have a ten digit number, abcdefghij. Each of the digits is different, and:

  • a is divisible by 1
  • ab is divisible by 2
  • abc is divisible by 3
  • abcd is divisible by 4
  • abcde is divisible by 5
  • abcdef is divisible by 6
  • abcdefg is divisible by 7
  • abcdefgh is divisible by 8
  • abcdefghi is divisible by 9
  • abcdefghij is divisible by 10

What’s my number?

[To clarify: a, b, c, d, e, f, g, h, i, and j are all single digits. Each digit from 0 to 9 is represented by exactly one letter. The number abcdefghij is a ten-digit number whose first digit is a, second digit is b, and so on. It does not mean that you multiply a x b x c x…]

And here are the hints we offered to help:

-If you add all the digits in a number, and the total is divisible by 3, then that number is also divisible by 3.
-If the last two digits of a number are divisible by 4, then that number is divisible by 4.
-If the last three digits of a number are divisible by 8, then that number is divisible by 8.


The solution is 3816547290.

So, how do we get there?

First, we use process of elimination.

Any number divisible by 10 must end in a zero, so j = 0.

Any number divisible by 5 must end in a zero or a five, so e = 5 (because each digit only appears once).

That gives us abcd5fghi0.

But that’s not all we know.

If a number is divisible by an even number, that number must itself be even. So that means b, d, f, and h must all be even numbers (i.e. some combination of 2, 4, 6, and 8). That also means that a, c, g, and i must all be some combination of the remaining odd numbers (1, 3, 7, and 9).

That’s a lot of information that will come in handy as we solve.

So, where to next? Let’s look at one of those even-numbered spots.

We’ve been told that abcd is divisible by 4. But any number is divisible by 4 if the last two digits are divisible by 4. So that means cd is divisible by 4.

So, if c is odd, d is even, and cd is divisible by 4, that limits the possibilities somewhat. cd must be 12, 16, 32, 36, 72, 76, 92, or 96.

So d is either 2 or 6.

That will be helpful in figuring out def. And knowing def is the key to this entire puzzle.


One of the clues we offered in our last post was that if the sum of a number’s digits is divisible by 3, then that number is also divisible by three. We know abc is divisible by 3, so that means a + b + c is also divisible by 3.

And if something is divisible by 6, then it’s also divisible by 3, so a + b + c + d + e + f is divisible by 3.

Here’s where things get a little tricky. Since a + b + c + d + e + f is divisible by 3, and a + b + c is divisible by 3, then when you subtract a + b + c from a + b + c + d + e + f, the result, d + e + f would also be divisible by 3.

Why is that helpful? Because it means we can look at def instead of abcdef, and we know a lot about def right now.

d is either 2 or 6. e is 5. f is either 2, 4, 6, or 8. And the sum of d + e + f is divisible by 3.

So that gives us two possibilities to deal with, either 2 + 5 + f, where the sum is divisible by 3, or 6 + 5 + f, where the sum is divisible by 3.

Since each number is only used once, that’s six possible equations:

  • 2 + 5 + 4 = 11
  • 2 + 5 + 6 = 13
  • 2 + 5 + 8 = 15
  • 6 + 5 + 2 = 13
  • 6 + 5 + 4 = 15
  • 6 + 5 + 8 = 19

Only 258 and 654 have sums divisible by 3, so they’re our two possibilities for def.

We’ll have to try both of them to see which is the correct choice. How do we do that?

Let’s start with the assumption that def is 258.


That would mean our answer is abc258ghi0. We know b and h have to be even numbers, and only 4 and 6 are left as options. Since fewer numbers are divisible by 8 than by 2, let’s look at abc258gh.

One of the other hints we offered was that if the last three digits of a number are divisible by 8, then the whole number is divisible by 8.

So that means if abc258gh is divisible by 8, then 8gh is divisible by 8. That’s much more manageable.

So, f is 8, h is 4 or 6, and g is either 1, 3, 7, or 9. That gives us eight possibilities for 8gh: 814, 834, 874, 894, 816, 836, 876, and 896.

Dividing each of these by 8 reveals only two possible choices: 816 and 896. That means, in this scenario, h is 6, b is 4, and our number is a4c258g6i0.

What’s next? Well, remember that trick we did with abcdef before? We’re going to do it again with abcdefghi.

Any number divisible by 9 is divisible by 3. Our rule of sums tells us that a + b + c + d + e + f + g + h + i is also divisible by 3. And since a + b + c + d + e + f is divisible by 3, subtracting it means that g + h + i is also divisible by 3.

With 816 and 896 as our possibilities for fgh, that means our possibilities for ghi are 16i and 96i. That gives us the following possibilities: 163, 167, 169, 961, 963, 967, where the sum of our answer must be divisible by 3.

  • 1 + 6 + 3 = 10
  • 1 + 6 + 7 = 14
  • 1 + 6 + 9 = 16
  • 9 + 6 + 1 = 16
  • 9 + 6 + 3 = 18
  • 9 + 6 + 7 = 22

963 is the only one that works, which gives us a4c2589630. With only 1 and 7 remaining as options, our possible solution is either 1472589630 or 7412589630.

But, if you divide either 1472589 or 7412589 by 7 — which is faster than running every one of the 10 conditions through a calculator — neither divides cleanly. That means 258 is incorrect.


I know that was a lot of work just to eliminate one possibility, but it was worth it. It means 654 is correct, so our solution so far reads abc654ghi0.

And we can use the same techniques we just employed with 258 to find the actual answer.

We know b and h have to be even numbers, and only 2 and 8 are left as options. Again, since fewer numbers are divisible by 8 than by 2, let’s look at abc654gh.

4gh is divisible is 8. So, f is 4, h is 2 or 8, and g is either 1, 3, 7, or 9. That gives us eight possibilities for 4gh: 412, 432, 472, 492, 418, 438, 478, and 498.

Dividing each of these by 8 reveals only two possible choices: 432 and 472. That means b is 8, and our number is a8c654g2i0.

Now, let’s look at ghi.

With 432 and 472 as our possibilities for fgh, that means our possibilities for ghi are 32i and 72i. That gives us the following possibilities: 321, 327, 329, 721, 723, 729, where the sum of our answer must be divisible by 3.

  • 3 + 2 + 1 = 6
  • 3 + 2 + 7 = 12
  • 3 + 2 + 9 = 14
  • 7 + 2 + 1 = 10
  • 7 + 2 + 3 = 12
  • 7 + 2 + 9 = 18

Okay, that leaves us four possibilities for ghi: 321, 327, 723, and 729.

Stay with me, folks, we’re so close to the end!

Let’s look at our four possibilities:

  • a8c6543210 (79)
  • a8c6543270 (19)
  • a8c6547230 (19)
  • a8c6547290 (13)

Next to each number, I’ve placed the only digits missing in each scenario, two for each.

That means there are only 8 possible ways to arrange the remaining numbers:

  • 7896543210
  • 9876543210
  • 1896543270
  • 9816543270
  • 1896547230
  • 9816547230
  • 1836547290
  • 3816547290

So let’s do what we did last time, and divide each chain at the seventh number by 7.

  • 7896543 / 7
  • 9876543 / 7
  • 1896543 / 7
  • 9816543 / 7
  • 1896547 / 7
  • 9816547 / 7
  • 1836547 / 7
  • 3816547 / 7

Only one of the chains can be cleanly divided by 7, and it’s 3816547.

Which means the solution for abcdefghij is 3816547290.


I know this was a monster of a solve — it rivals our Brooklyn Nine-Nine seesaw puzzle solution in complexity — but it’s one that every one of our fellow PuzzleNationers are capable of puzzling out.

How did you do on this diabolical brain teaser, folks? Let us know in the comments section below. We’d love to hear from you!


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1 thought on “Answer to the Fiendish Second Conway Puzzle, The Ten Divisibilities!

  1. Just after d = 2 or 6 you can get that h = 2 or 6 also by using that in the even hundreds, multiples of 8 with a leading odd and non-zero units are 16,32,56,72,96 implying that h = 2 or 6. So d and h cover 2 and 6 meaning b and f must be 4 and 8. This neatens the very clever multiple of three arguments to follow. Great puzzle..

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