2006 AMC 8 Problems/Problem 23: Difference between revisions
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==Solution== | ==Solution== | ||
(A) The counting numbers that leave a remainder of 4 when divided by 6 are | ('''A''') The counting numbers that leave a remainder of 4 when divided by 6 are | ||
4; 10; 16; 22; 28; 34;...; etc. The counting numbers that leave a remainder of 3 when | 4; 10; 16; 22; 28; 34;...; etc. The counting numbers that leave a remainder of 3 when | ||
divided by 5 are 3; 8; 13; 18; 23; 28; 33;...; etc. So 28 is the smallest possible number | divided by 5 are 3; 8; 13; 18; 23; 28; 33;...; etc. So 28 is the smallest possible number | ||
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by both 6 and 5. The smallest number that is divisible by 6 and 5 is 30, so the | by both 6 and 5. The smallest number that is divisible by 6 and 5 is 30, so the | ||
smallest possible number of coins in the box is 28. | smallest possible number of coins in the box is 28. | ||
{{AMC8 box|year=2006|n=II|num-b=22|num-a=24}} | |||
Revision as of 13:53, 3 November 2012
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 ¯ve 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? (A) 0 (B) 1 (C) 2 (D) 3 (E) 5
Solution
(A) The counting numbers that leave a remainder of 4 when divided by 6 are 4; 10; 16; 22; 28; 34;...; etc. The counting numbers that leave a remainder of 3 when divided by 5 are 3; 8; 13; 18; 23; 28; 33;...; etc. So 28 is the smallest possible number of coins that meets both conditions. Because 4 £ 7 = 28, there are no coins left when they are divided among seven people.
OR
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 30, so the smallest possible number of coins in the box is 28.
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
