2006
Problem - 991
A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?
Answer
A
We note that the remainder is always two less than the divisor when the coins are distributed among six or five people. Therefore, the number of coin must be two less than the least common multiple of $6$ and $5$, which is $28$. It follows the answer is $\boxed{0}$.