Final answer:
If $n$ is a nonnegative integer and the sum of its digits is divisible by 3, then $n$ is divisible by 3. This is shown using the property that a number is divisible by 3 if its digit sum is divisible by 3. By expressing $n$ as the sum of its digits times powers of 10 and examining the divisibility by 3, the result is confirmed.
Step-by-step explanation:
To prove that if $n$ is a nonnegative integer such that the sum of its digits is divisible by 3, then 3 divides $n$, we can use the property of divisibility by 3. According to this property, an integer is divisible by 3 if and only if the sum of its digits is divisible by 3. Let's consider a nonnegative integer $n$ with digits $d_kd_{k-1}...d_2d_1d_0$ where $d_k$ is the most significant digit and $d_0$ is the least significant digit. The value of $n$ can be expressed as: $n = d_k \times 10^k + d_{k-1} \times 10^{k-1} + ... + d_2 \times 10^2 + d_1 \times 10 + d_0$ Notice that $10^k$, $10^{k-1}$, ..., $10^2$, and $10$ are all powers of 10, and each power of 10 is divisible by 3 except for a remainder of 1. Therefore, when we divide each term except the last digit by 3, the remainders will cancel each other out. The sum of the digits of $n$ is $d_k + d_{k-1} + ... + d_2 + d_1 + d_0$ which is given to be divisible by 3. Since the sum of the digits is divisible by 3, it implies that the whole number $n$ must also be divisible by 3. Thus, we have shown that if the sum of the digits of a nonnegative integer $n$ is divisible by 3, then 3 divides $n$.