2006 AMC 8 Problems/Problem 23: Difference between revisions
Firestarly (talk | contribs) 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..." |
Firestarly (talk | contribs) |
||
| Line 11: | Line 11: | ||
(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; | 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; | 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.
