# 2005 AMC 10B Problems/Problem 13

## Problem

How many numbers between $1$ and $2005$ are integer multiples of $3$ or $4$ but not $12$? $\textbf{(A) } 501 \qquad \textbf{(B) } 668 \qquad \textbf{(C) } 835 \qquad \textbf{(D) } 1002 \qquad \textbf{(E) } 1169$

## 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 $668+501-167-167 = \boxed{\textbf{(C) } 835}$

## 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 $\frac{5}{12} \cdot 2005$ = $\boxed{\textbf{(C) }835}$

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