Difference between revisions of "2005 AMC 10B Problems/Problem 13"
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== Solution 1 == | == 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 | + | To find the multiples of <math>3</math> or <math>4</math> but not <math>12</math>, you need to find the number of multiples of <math>3</math> and <math>4</math>, and then subtract twice the number of multiples of <math>12</math>, because you overcount and do not want to include them. The multiples of <math>3</math> are <math>\frac{2005}{3} = 668\text{ }R1.</math> The multiples of <math>4</math> are <math>\frac{2005}{4} = 501 \text{ }R1</math>. The multiples of <math>12</math> are <math>\frac{2005}{12} = 167\text{ }R1.</math> So, the answer is <math>668+501-167-167 = \boxed{\mathrm{(C)}\ 835}</math> |
== Solution 2 == | == Solution 2 == |
Revision as of 21:37, 4 July 2019
Contents
[hide]Problem
How many numbers between and are integer multiples of or but not ?
Solution 1
To find the multiples of or but not , you need to find the number of multiples of and , and then subtract twice the number of multiples of , because you overcount and do not want to include them. The multiples of are The multiples of are . The multiples of are So, the answer is
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 |
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