Answer
See explanation
Work Step by Step
Let \( n \) be any nonnegative integer with decimal representation
\[
n = d_k 10^k + d_{k-1} 10^{k-1} + \cdots + d_1 10 + d_0,
\]
where each \( d_i \) is a digit (between 0 and 9). Notice that
\[
10^i = (9 + 1)^i,
\]
and by the binomial theorem every term in the expansion of \( (9+1)^i \) except the last is divisible by 9. More directly, since
\[
10 \equiv 1 \pmod{9},
\]
it follows that
\[
10^i \equiv 1^i \equiv 1 \pmod{9}
\]
for every nonnegative integer \( i \). Thus, we have
\[
n \equiv d_k + d_{k-1} + \cdots + d_1 + d_0 \pmod{9}.
\]
This means that \( n \) and the sum of its digits have the same remainder when divided by 9. Therefore, if the sum of the digits of \( n \) is divisible by 9 (i.e. the sum is congruent to 0 modulo 9), then \( n \) must also be divisible by 9.
This completes the proof.
