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, 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 (ProblemsAnswer KeyResources)
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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