We have a 8 figure number, lets call it $N_1=a_1a_2a_3a_4a_5a_6a_7a_8$. Prove that if we interchange two numbers (for example $N_2=a_1a_2a_4a_3a_5a_6a_7a_8$ and $N_1 \neq N_2$) then the rest we get by dividing $N_1$ and $N_2$ with 23 isn't the same. Then prove that if we change one number (for example: $N_3=a_1a_2ba_4a_5a_6a_7a_8$ and $b\neq a_3$) the rest we get by dividing $N_1$ and $N_3$ by 23 is not the same.

Then prove that this is not true if we divide the numbers by 24. I mean, that we can find $N_1, N_2$ and $N_3$ defined as I said before that after dividing by 24 the rest is the same.

I do not even know how to start. Thanks in advance.

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    $\begingroup$ Well, the first claim isn't true as stated...since nothing prevents $a_3=a_4$. If you add the condition that $a_i\neq a_j$, then $N_1-N_2=(a_i-a_j)\times 10^{8-i}-(a_j-a_i)10^{8-j}$. (Note: check the exponents there). Work from there. $\endgroup$ – lulu Nov 25 '18 at 14:04
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    $\begingroup$ Side note: life (or at least arithmetic) gets easier if you change notation to $N_1=a_7a_6\cdots a_0=\sum a_i10^i$. $\endgroup$ – lulu Nov 25 '18 at 14:12
  • $\begingroup$ @lulu I edited it. Not all the figures have to be different, but the new number $N_2$ can't be the same as $N_1$ (so the interchanged figures can not be the same figure). $\endgroup$ – Andarrkor Nov 25 '18 at 14:49
  • $\begingroup$ Sure. As I said, I figured you meant to add that condition. $\endgroup$ – lulu Nov 25 '18 at 15:02
  • $\begingroup$ @lulu I do not know what that subtraction has to do with the rest. I don't understand why this changes when the dividing number is 24. I suppose that it is something related to prime numbers, as 23 is prime. Thanks for the answers. $\endgroup$ – Andarrkor Nov 25 '18 at 15:09

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