My attempt:

If \(N^3\) ends in 10, that means it is divisible by 10. That means that both 2 and 5 divide \(N^3\). If we decompose \(N\) into its prime factors \(N = {p_1}^{e_1}{p_2}^{e_2} \dots {p_k}^{e_k}\), then \(N^3 = {p_1}^{3e_1}{p_2}^{3e_2} \dots {p_k}^{3e_k}\). We know that 2 and 5 have to divide \(N\), which implies that there are at least two other factors of 2 and two other factors of 5 that divide \(N^3\). That means that \(N^3\) is at least divisible by 1000, which would make it end in three zeros, not one. Thus we reach a contradiction, so \(N^3\) can't end in 10.

Edited October 15th, 2017
by EN