1966 AHSME Problems/Problem 29

Problem

The number of positive integers less than $1000$ divisible by neither $5$ nor $7$ is:

$\text{(A) } 688 \quad \text{(B) } 686 \quad \text{(C) } 684 \quad \text{(D) } 658 \quad \text{(E) } 630$

Solution

The number of positive integers under $1000$ that are divisible by $5$ is $\lfloor\frac{999}{5}\rfloor=199$. The number of positive integers under $1000$ that are divisible by $7$ is $\lfloor\frac{1000}{7}\rfloor=142$. Adding them together, we get $341$. However, we have over counted the numbers which are divisible by $35$. There are $\lfloor\frac{999}{35}\rfloor=28$ of these. So, the number of positive integers divisible be $7$ or $5$ under $1000$ is $341-28=313$. We can conclude that the number of positive integers divisible by neither $5$ or $7$ is $999-313=686$ or answer choice $\fbox{B}$.

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
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