Problem:
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 Choices:
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Solution:
The counting numbers that leave a remainder of when divided by are The counting numbers that leave a remainder of when divided by are So is the smallest possible number of coins that meets both conditions. Because , there are no coins left when they are divided among seven people.
If there were two more coins in the box, the number of coins would be divisible by both and . The smallest number that is divisible by and is , so the smallest possible number of coins in the box is .
Answer: .
The problems on this page are the property of the MAA's American Mathematics Competitions