Here is the problem:

A nine-digit integer is called beautiful if all of its digits are different. Prove that there exist at least $1000$ beautiful numbers, each of which is divisible by $37.$

Here is what I did:

Let $n$ be a beautiful number which is divisible by $37$ ,we have $$n=\overline{a_9a_8\ldots a_1}$$ Otherwise , $10^3\equiv 1 \mod 37$

so $$10^2(a_9+a_6+a_3)+10(a_8+a_5+a_2)+(a_7+a_4+a_1)\equiv 0 \mod 37$$

$$-11(a_9+a_6+a_3)+10(a_8+a_5+a_2)+(a_7+a_4+a_1)\equiv 0 \mod 37$$ $$10(a_8+a_5+a_2-a_9-a_6-a_3)+(a_7+a_4+a_1-a_9-a_6-a_3)\equiv 0 \mod 37$$ we see that if $a_8+a_5+a_2=a_9+a_6+a_3= a_7+a_4+a_1$ we would be done

Otherwise $1+9+5=3+4+8=6+7+2$ so $a_8$ has 9 choices , $a_5$ has $2$ , $a_9$ has 6 , $a_6$ has $2$ , $a_7$ has $3$ , $a_4$ has $2$

So there are at least $1296$ beautiful numbers each of which is divisible by $37$

I just want to know if this is right.

  • $\begingroup$ Can you link to a source of this (nice) problem? $\endgroup$ – coffeemath Mar 9 at 22:52
  • $\begingroup$ maths.usyd.edu.au/u/dzmitry/TT_problems/TT40JO.pdf $\endgroup$ – user600785 Mar 10 at 14:16
  • $\begingroup$ In the two places where you say "Otherwise," I think you mean "Because," and where you say "I just want if this is right," I'm pretty sure you mean "I just want to know if this is right." But where you say "unless don't," do you want a correct solution if yours is not, or do you just want to know if yours is correct? $\endgroup$ – Barry Cipra Mar 10 at 14:45
  • $\begingroup$ thanks , i don't want a correct solution , i just want to know if this is right $\endgroup$ – user600785 Mar 10 at 15:32

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