Let the decimal representation of \(n\) be \[n = d_k d_{k-1} \dots d_2 d_1\] Expand the number at the one's place. \[d_k d_{k-1} \dots d_2 \cdot 10 + d_1\] \[= 10 \cdot k + d_1\] where \(k\) represents the integer whose decimal expansion is \(d_k d_k-1 \dots d_2.\)

If \(d_1\) is a multiple of \(10\) then \(d_1 = 10 \cdot j\) for some integer \(j,\) in which case \(n = 10 \cdot (k + j)\) and is divisible by \(10.\) On the other hand, if \(d_1\) is not divisible by \(10\) then \(d_1 = 10 \cdot j + r\) for some integers \(j\) and \(1 \leq r \leq 9,\) in which case \(n = 10 \cdot (k + j) + r\) is not divisible by \(10.\)

Since \(d_1\) is a digit, \(0 \leq d_1 \leq 9.\) Therefore, \(d_1\) is only divisible by \(10\) if \(d_1 = 0\). So, \(n\) is divisible by \(10\) if and only if \(n\) has a \(0\) in the one's place.