2006 AMC 8 Problems/Problem 23
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?
Solution
Solution 1
The counting numbers that leave a remainder of 4 when divided by 6 are
The counting numbers that leave a remainder of 3 when
divided by 5 are
So 28 is the smallest possible number
of coins that meets both conditions. Because
, there are
coins left
when they are divided among seven people.
Solution 2
If there were two more coins in the box, the number of coins would be divisible
by both 6 and 5. The smallest number that is divisible by 6 and 5 is , so the
smallest possible number of coins in the box is
and the remainder when divided by 7 is
.
See Also
2006 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
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