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Difference between revisions of "2005 AMC 10B Problems/Problem 13"

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== Solution 1 ==
 
== Solution 1 ==
  
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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>
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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

Problem

How many numbers between $1$ and $2005$ are integer multiples of $3$ or $4$ but not $12$?

$\mathrm{(A)} 501 \qquad \mathrm{(B)} 668 \qquad \mathrm{(C)} 835 \qquad \mathrm{(D)} 1002 \qquad \mathrm{(E)} 1169$


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 $\frac{2005}{3} = 668\text{ }R1.$ The multiples of $4$ are $\frac{2005}{4} = 501 \text{ }R1$. The multiples of $12$ are $\frac{2005}{12} = 167\text{ }R1.$ So, the answer is $668+501-167-167 = \boxed{\mathrm{(C)}\ 835}$

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 $\frac{5}{12}$ of positive integers are multiples of 3 or 4 but not 12, the answer is approximately $\frac{5}{12} \cdot 2005$ = $\boxed{\mathrm{(C)}\ 835}$

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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