Let M and N be single-digit integers. If the product 2M5 x 13N is divisible by 36, how many ordered pairs (M,N) are possible?

-- source

I tried it by reducing 36 into its positive factors (1,2,3,4,6,9,18,36) and then solving, but I got way too many pairs. Can somebody help?

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    $\begingroup$ What do you mean by $2M5$ and $13N$? Are these concatenated numbers (two hundred something and five, one hundred and thirty something) or products (two times M times five, thirteen times N)? $\endgroup$ Commented Jun 6, 2019 at 9:32
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    $\begingroup$ Also, what is your source for these questions? $\endgroup$ Commented Jun 6, 2019 at 9:33
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    $\begingroup$ Or power $2M^5$? $\endgroup$ Commented Jun 6, 2019 at 9:38
  • $\begingroup$ To the close-voters: as the first answerer here, I think this is mathematically "interesting" enough not to be closed. It requires some knowledge of basic number theory (prime factorisations) if you want to do it in a quick neat way rather than brute-forcing all possibilities. $\endgroup$ Commented Jun 6, 2019 at 15:36
  • $\begingroup$ Sorry for answering so late. Yes, @Randal'Thor, 2M5 and 13N are concatenated numbers. And as for the source of these questions, I have a book. $\endgroup$ Commented Jun 6, 2019 at 16:37

2 Answers 2


I'll assume here that the number $2M5$ and $13N$ are concatenated three-digit numbers rather than products, because if they are products then the question is trivial.

Being divisible by 36 is equivalent to

being divisible by both of its prime factors $4=2^2$ and $9=3^2$.

  1. $2M5$ is odd, so both factors of 2 must come from $13N$. That means $3N$ must be a multiple of 4, so either $N=2$ or $N=6$. In either of these cases, the product is definitely a multiple of 4 - it's an "if and only if" condition.

  2. If the product is divisible by 9, then either both factors are multiples of 3 (i.e. their digit sums are multiples of 3) or one of the two factors is a multiple of 9 (i.e. its digit sum is a multiple of 9).

    Let's consider the two cases from the first numbered point above:

    • If $N=2$, then $132$ is divisible by 3 but not 9, so we need $2M5$ divisible by 3, i.e. $2+M+5$ divisible by 3, which means either $M=2$ or $M=5$ or $M=8$. In all of these cases, the product is definitely a multiple of 9.

    • If $N=6$, then $136$ is not divisible by 3, so we need $2M5$ divisible by 9, i.e. $2+M+5$ divisible by 9, which means $M=2$. Again, in this case the product is definitely a multiple of 9.

So the possibilities for the pair $(M,N)$ are:

$(2,2),(5,2),(8,2),(2,6)$ - four possibilities in all.

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    $\begingroup$ @WeatherVane It's nothing to do with "you" or "me" - don't take it personally. I don't like answers which use computers instead of elegant mathematics, but i don't go around downvoting or deleting them either (unless they're on questions tagged no-computers, which this isn't). On this site, we're commenting/voting on content not people, as you should know :-) $\endgroup$ Commented Jun 6, 2019 at 10:55
  • $\begingroup$ Just to balance the comment you decided to leave, nobody downvoted or deleted a question, or suggested the mutual voting you accused me of requesting. You could have deleted it with the others, but for some reason decided not to. $\endgroup$ Commented Jun 6, 2019 at 17:42

Confirmation of the answer from Rand al'Thor using brute force with C - only 100 perms:

 #include <stdio.h>
 int main(void)
     for(int M = 0; M < 10; M++)
         for(int N = 0; N < 10; N++)
             if(((205 + 10 * M) * (130 + N)) % 36 == 0)
                 printf("%d %d\n", M, N);
2 2
2 6
5 2
8 2


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