2005 AMC 10B Problems/Problem 13
Contents
[hide]Problem
How many numbers between and are integer multiples of or but not ?
Solution 1
To find the multiples of 3 or 4 but not 12, you need to find the number of multiples of 3 and 4, and then subtract twice the number of multiples of 12, because you overcount and do not want to include them. The multiples of 3 are 2005/3 = 668 R1. The multiples of 4 are 2005/4 = 501 R1. The multiples of 12 are 2005/12 = 167 R1. So, the answer is 668+501-167-167 =
Solution 2
From 1-12, the multiples of 3 or 4 but not 12 are 3, 4, 6, 8, an 9, a total of five numbers. Since of positive integers are multiples of 3 or 4 but not 12, the answer is approximately =
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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All AMC 10 Problems and Solutions |
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