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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? (2006 AMC 8 Problem, Question \#23) 
Overseas Competition->Knowledge Point->Number Theory Modules
The counting numbers that leave a remainder of 4 when divided by 6 are $4,10,16,22,28,34, \cdots$ The counting numbers that leave a remainder of 3 when divided by 5 are $3,8,13,18,23,28,33, \cdots$ So 28 is the smallest possible number of coins that meets both conditions. Because $4 \cdot 7=28$, there are (A) 0 coins left when they are divided among seven people. 