2006 AMC 8 Problem 23

Below is the professionally curated solution for Problem 23 of the 2006 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 8 solutions, or check the answer key.

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Concepts:modular arithmeticsystematic listing

Difficulty rating: 1450

23.

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?

00

11

22

33

55

Solution:

The positive integers that leave a remainder of 44 when divided by 66 are 4,10,16,22,28,34,40,. 4, 10, 16, 22, 28, 34, 40, \cdots. The positive integers that leave a remainder of 33 when divided by 55 are 3,8,13,18,23,28,33,. 3, 8, 13, 18, 23, 28, 33, \cdots.

From this, we can see that the smallest number of coins that work is 28.28. This leaves a remainder of 00 when divided by 7.7.

Thus, A is the correct answer.

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