Difference between revisions of "2005 AMC 10B Problems/Problem 13"
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| − | == Solution == | + | == 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 = <math>\boxed{\mathrm{(C)}\ 835}</math> | 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 = <math>\boxed{\mathrm{(C)}\ 835}</math> | ||
| + | |||
| + | == 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 <math>\frac{5}{12}</math> of positive integers are multiples of 3 or 4 but not 12, the answer is approximately <math>\frac{5}{12} \cdot 2005</math> = <math>\boxed{\mathrm{(C)}\ 835}</math> | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=12|num-a=14}} | {{AMC10 box|year=2005|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:23, 4 February 2019
Contents
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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
