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 numbers under $1000$ that are divisible by $5$ is $\lfloor\frac{999}{5}\rfloor=199$ . The number of numbers 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 numbers divisible be $2$ or $5$ under $1000$ is $341-28=313$ . We can conclude that the number of numbers divisible by neither $5$ or $7$ is $999-313=686$ or answer choice $\fbox{B}$ .

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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 • 26 • 27 • 28 • 29 • 30
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1966 AHSME Problems/Problem 29

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