# How to prove that if a figure in a number changes, then the rest also changes.

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.

• 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. – lulu Nov 25 '18 at 14:04
• Side note: life (or at least arithmetic) gets easier if you change notation to $N_1=a_7a_6\cdots a_0=\sum a_i10^i$. – lulu Nov 25 '18 at 14:12
• @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). – Andarrkor Nov 25 '18 at 14:49
• Sure. As I said, I figured you meant to add that condition. – lulu Nov 25 '18 at 15:02
• @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. – Andarrkor Nov 25 '18 at 15:09