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I'm confused about this and I don't know how to prove it. I tried to prove it but I want to know if I'm in the right way or not.

My prove:

The statement $10 \equiv 1$ (mod 9) means that the remainder when 10 is divided by 9 is 1. We can use this statement to prove the divisibility rule for 9.

If an integer is divisible by 9, then the sum of its digits is also divisible by 9. To see this, let the integer be expressed in base 10 as $a_na_{n-1}...a_1a_0$, where each $a_i$ is a digit. Then we have

$a_n*10^n + a_{n-1}10^{n-1} +... + a_110 + a_0 \equiv a_n + a_{n-1} +... + a_1 + a_0$ (mod 9)

since each term $a_i*10^i$ is congruent to 0 (mod 9). So the integer is divisible by 9 if and only if its digits add up to a multiple of 9.

To prove the other way, if the sum of the digits of an integer is divisible by 9. Then the integer is divisible by 9. we can simplify the above congruence expression:

$a_n + a_{n-1} +... + a_1 + a_0 \equiv a_n*10^n + a_{n-1}10^{n-1} +... + a_110 + a_0$ (mod 9)

the LHS is the sum of digits which is divisible by 9 and the RHS is the original number, thus the original number is also divisible by 9.

Bill Dubuque
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Amirreza Hashemi
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  • It is not true that "each term $a_i\times 10^i$ is congruent to $0\pmod 9$". Indeed, that term is congruent to $a_i\pmod 9$. – lulu Jan 11 '23 at 20:23
  • This: "since each term $_ 10^i$ is congruent to 0 (mod 9)" is wrong. $10^1$ is congruent to 1 mod 9. – Ethan Bolker Jan 11 '23 at 20:23
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    The proof is employs basic congruence laws, notably the congruence Sum, Product & Power rules (or the more general Polynomial congruence rule), as explained here in the linked dupe. By $10\equiv 1$ we get $10^i\equiv 1^i\equiv 1$ by the Power rule so $,a 10^i\equiv a\cdot 1\equiv a,$ by the Product rule. Or, simpler by the Polynomial rule: $10\equiv 1\Rightarrow f(10)\equiv f(1) \equiv $ digit sum, where $f(x)$ is your displayed radix $10$ polynomial for the given integer. – Bill Dubuque Jan 11 '23 at 20:30

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