# 2005 AMC 10B Problems/Problem 13

## 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, and 9, a total of five numbers. Since $\frac{5}{12}$ of positive integers are multiples of 3 or 4 but not 12 from 1-12, the answer is approximately $\frac{5}{12} \cdot 2005$ = $\boxed{\mathrm{(C)}\ 835}$

## See Also

 2005 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 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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