Difference between revisions of "2006 AMC 8 Problems/Problem 23"

(Created page with "==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 coi...")
 
(Solution)
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(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; : : :. 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; : : :. 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
 
of coins that meets both conditions. Because 4 £ 7 = 28, there are no coins left
 
of coins that meets both conditions. Because 4 £ 7 = 28, there are no coins left
 
when they are divided among seven people.
 
when they are divided among seven people.
 +
 
OR
 
OR
 +
 
If there were two more coins in the box, the number of coins would be divisible
 
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
 
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.

Revision as of 18:27, 31 October 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.