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How to prove that :

9 | $a_k+... +a_1+a_0$ implies that

9 | $a_k 10^k +... +a_1 10^1 + a_o 10^0$

where | denotes the divisibility sign (that is, I need to prove that if $a_k+... a_1+a_0$ is divisible by 9 then so is $a_k 10^k +... a_1 10^1 + a_o 10^0$).

Tony
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  • It is actually a co-implication. –  Nov 17 '16 at 02:09
  • It is sufficient to show the difference is divisible by 9. It is sufficient to show (10^k-1) is divisible by 9. But this is 10^k-1^k, which by the formula for the difference of kth powers is divisible by 9. – Jacob Wakem Nov 17 '16 at 02:25

4 Answers4

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Note that $10 \equiv 1 \pmod{9}$ so $10^n \equiv 1^n \equiv 1 \pmod{9}$. This means $$a_k+ \cdots + a_0 \equiv a_k10^k+ \cdots + a_0 \pmod{9}.$$

Tengu
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Hint: Apply the binomial expansion to $10^n=(9+1)^n$.

Reveillark
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Expand $(9+1)^j$. In the expansion, every term has a factor $9$ except for $1$.That's where you are to apply your hypothesis.

Isko10986
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For a number %n% to be divisible by 9 the sum of it's digit must be divisible by 9. So, $a_k 10^k +… + a_1 10^1 + a_o 10^0$ holds true if, for the k digit number n $$S = a_k +… + a_1 + a_o$$ is divisible by 9, i.e. $9 | a_k+... +a_1+a_0$ which is the provided statement.

Inquistador
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