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 \( d_0 \) is the rightmost digit, \( d_1 \) the next digit to the left, and so on. Notice that
\[
10 \equiv -1 \pmod{11}.
\]
Raising both sides to the \( i \)th power, we have
\[
10^i \equiv (-1)^i \pmod{11}.
\]
Thus, we can rewrite \( n \) modulo 11 as
\[
n \equiv d_k(-1)^k + d_{k-1}(-1)^{k-1} + \cdots + d_1(-1)^1 + d_0(-1)^0 \pmod{11}.
\]
This expression is exactly the alternating sum of the digits of \( n \) (starting from the rightmost digit \( d_0 \) as positive, then subtracting \( d_1 \), adding \( d_2 \), etc.). In other words,
\[
n \equiv \text{(alternating sum of the digits of } n\text{)} \pmod{11}.
\]
Therefore, if the alternating sum of the digits of \( n \) is divisible by 11 (i.e. it is congruent to 0 modulo 11), then
\[
n \equiv 0 \pmod{11},
\]
which means that \( n \) is divisible by 11.
This completes the justification.
