If you take a problem of the form:


where each letter must be replaced by a different digit (that is, all digits, 0-9 must be used once), how many answers are there? I know of at least a few, but is there a way to find all of them without brute forcing the problem?

One possible answer is:


can you find all the others?

Clarification: No leading zeros, please! (i.e. g, a, and d must not be 0)

  • $\begingroup$ Leading zeroes allowed? $\endgroup$ Commented May 23, 2017 at 19:44
  • 3
    $\begingroup$ I used a computer, and there are 96 possible answers. I think this may be a bit broad $\endgroup$ Commented May 23, 2017 at 19:46
  • $\begingroup$ good question @BeastlyGerbil , I'm gonna go ahead and say no leading zeros. And it might be too broad, hmmm. I was just wondering if there was a good algorithm for finding all of them without brute forcing it, but if there are that many maybe there isn't? $\endgroup$
    – MMAdams
    Commented May 23, 2017 at 20:00
  • $\begingroup$ a priori, why would there be a reason for an algorithm to exist? $\endgroup$
    – Wen1now
    Commented May 23, 2017 at 21:16

2 Answers 2


Partial answer / possible strategy.

Let's say we have a solutions.
Just by interchanging c with f we get an other solution.
The same goes if we change b and e or a and d.
This means that the total number of solutions is a multiple of $2 \times 2 \times 2 = 8$

now finding the solutions:

A place to start would be to group 1 and 0 to form 10 and look for combinations in the numbers 2 to 10 (9 numbers) in such a way that 2 numbers add up to a third one. So we don't get any carry over except in the hundreds place.
This way we get at least 2 solutions (without interchanging digits between numbers) for a combination because since there is no carry over we can switch the tenths with the units (inside both the numbers in the addition)
Such a solutions would be:
$5 + 4 = 9$, $2+6 = 8$ and $7+3 = 10$
So we get the solutions: $725 + 364 = 1089$ and $752 + 346 = 1098$.
based on the logic above we now have $2 \times 8 = 16$ solutions.
An other combination would be:
$3+5 = 8$, $7+2 = 9$ and $4+6 = 10$.
So we get the solutions:
$473 + 625 = 1098$ and $437 + 652 = 1089$ and the additional solutions obtained from interchanging values (16 in total).
So now we have at least $16 + 16 = 32$ solutions.

Next step.

g is clearly 1 since we can get a carry over of max 1 from the hundreds place.
So we are left with the digits 0 and 2 to 9 (9 digits) to group them in such a way that at least 2 of them (but not just the ones we place in the hundreds place) add up over 10 so we get a carry over so we can get different solutions from the case above.

One possible way of doing so:

$9+4 = 13$ (units place), $6+8 = 14$ (tenths place and we get a carry over from the units place so we end up with 15) and $7+2 = 9$ (on the hundreds place and we get the carry over from the tenths to reach 10).
The solutions is:
$769 + 284 = 1053$ and by interchanging the digits we get 8 solutions in total. Now we have 40.

This can go on until we find all possible combinations.

Indeed using a computer (sorry for that, but I got tired of blindly stabbing for solutions) you get 96 total solutions. Which matches my finding that the total number is a multiple of 8.

  • $\begingroup$ This is an extremely well thought out answer! Thank you! So, I guess there's kind of 12 "unique" answers and then 8 permutations of those then. $\endgroup$
    – MMAdams
    Commented May 24, 2017 at 13:32
  • $\begingroup$ I guess so. The math does not lie. $\endgroup$
    – Marius
    Commented May 24, 2017 at 14:38

If A=7 B=8 C=9 D=2 E=4 F=6 then G=1 H=0 I=3 J=5


If A=7 B=8 C=6 D=2 E=4 F=9 then G=1 H=0 I=3 J=5


I just tried for these 2 answers, I think still there are many possibilities.

  • $\begingroup$ See my comment there are another 94 ways $\endgroup$ Commented May 23, 2017 at 20:53
  • $\begingroup$ Okay, Based on condition (No leading zeros), How can you know still 94. I don't know exact count I am just asking? $\endgroup$
    – CR241
    Commented May 23, 2017 at 21:01
  • $\begingroup$ Just try all permutations. There are 10 letters, so there are 10! possible permutaions. That's only about 3.6 million evaluations, which your computer can handle very quickly. $\endgroup$
    – M Oehm
    Commented May 24, 2017 at 5:31
  • $\begingroup$ @MOehm no-computers tag though $\endgroup$ Commented May 24, 2017 at 6:11
  • $\begingroup$ @BeastlyGerbil: I've sen the tag. That comment was a reply to the question in CR241's comment. I share your criticism that the question is too broad (and probably not very interesting) when the right thing to do [tm] is to throw computing power at it. $\endgroup$
    – M Oehm
    Commented May 24, 2017 at 6:19

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