2006 AMC 8 Problem 23
Attempt Problem 23 of the 2006 AMC 8 below, then check your answer against the professionally curated solution 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.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
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?
Answer: A
Small hint:
List small numbers that leave remainder when divided by .
Big hint:
Equivalently, adding two coins would make the number divisible by both and .
Solution:
The positive integers that leave a remainder of when divided by are The positive integers that leave a remainder of when divided by are
From this, we can see that the smallest number of coins that work is This leaves a remainder of when divided by
Thus, A is the correct answer.
Problem 23 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
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- 1989 AMC 8
- 1990 AMC 8
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