2002 AMC 8 Problem 2
Attempt Problem 2 of the 2002 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 2002 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
2.
How many different combinations of bills and bills can be used to make a total of Order does not matter in this problem.
Answer: A
Small hint:
Because is odd, the number of bills must be odd.
Big hint:
Try the possible odd numbers of bills before the total goes over
Solution:
Since the total is odd, the number of bills must be odd.
One bill leaves which is six bills. Three bills leave which is one bill. Five bills is too much.
Therefore, there are combinations.
Thus, A is the correct answer.
Problem 2 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
- 1994 AMC 8
- 1995 AMC 8
- 1996 AMC 8
- 1997 AMC 8
- 1998 AMC 8
- 1999 AMC 8
- 2000 AMC 8
- 2001 AMC 8
- 2003 AMC 8
- 2004 AMC 8
- 2005 AMC 8
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- 2007 AMC 8
- 2008 AMC 8
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