National Math Olympiad

Let $a_1$ denote the number of golden coins given to the first pirate, $a_2$ the number of golden coins given to the second pirate, and so on. Since $10\nmid a_i-a_j$ for all $i\ne j$, the numbers $a_1,a_2,\dots ,a_{10}$ must give different remainders when divided by $10$. There are only $10$ possible remainders, so these are exactly all the numbers $a_i$. Write $a_i=10k_i+l_i$ where $l_i$ is the remainder of $a_i$ when dividing by $10$. The total number of golden coins is equal to $$a_1+a_2+\cdots +a_{10}=10(k_1+\cdots +k_{10})+(l_1+\cdots +l_{10}).$$ Since the numbers $l_1,\dots ,l_{10}$ are exactly the numbers $0,1,2,\dots ,9$ (just not necessarily in this order), their sum is equal to $l_1+l_2+\cdots +l_{10}=0+1+2+\cdots +9=45$. We conclude that there were $10(k_1+\cdots +k_{10})+45$ golden coins in the chest.

Now, let us assume that the pirates can divide the silver coins in the same way. As above we come to the conclusion that the total number of silver coins is equal to $10(m_1+m_2+\cdots +m_{10})+45$ for some integers $m_1,\dots ,m_{10}$. But the total number of silver coins is even, a contradiction.