Question

Suppose n=(a_k)10^k+(a_k-1)10^(k-1)+....+(a_1)10+a_0
where 0<=a_j<=9 for all j.

Prove n==a_0+a_1+...+a_k(mod 9)

Answer 1

Suppose
\[n=a _{k}10^{k}+a _{k-1}10^{k-1}+...+a _{1}10+a _{0}\]
where
\[0 \le a _{j} \le 9\]

Prove the following:
\[n =a _{0}+a _{1}+....+a _{k} \left(\mod9 \right)\]

Answer 2

the only difference is that the equal sign is suppose to be the one with 3 bars instead of 2