Exercise 0.3.3 (Casting out nines)

Question

Prove that if $a = a_n 10^n + a_{n-1} 10^{n-1}+ \cdots + a_1 10 +a_0$ is any positive integer then $a \equiv a_n + a_{n-1} + \cdots + a_1 + a_0 \pmod 9$.

Richard Ganaye · Accepted Answer

\begin{proof} Since $10 \equiv 1 \pmod 9$, we obtain by Theorem 3 and induction that $10^k \equiv 1$ for every integer $k\geq 0$. Using Theorem 3 anew, 
$$a = \sum_{k=0}^n a_k 10^k \equiv \sum_{k=0}^n a_k,$$
so 
$$a \equiv a_n + a_{n-1} + \cdots + a_1 + a_0 \pmod 9.$$


\end{proof}

Exercise 0.3.3 (Casting out nines)

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Exercise 0.3.3 (Casting out nines)

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