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1991 AJHSME Problems/Problem 9

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Problem

How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

\text{(A)}\ 18 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 25 \qquad \text{(E)}\ 27

Solution

There are \left\lfloor \frac{46}{3}\right\rfloor =15 numbers divisible by 3, \left\lfloor\frac{46}{5}\right\rfloor =9 numbers divisible by 5, so at first we have 15+9=24 numbers that are divisible by 3 or 5, except we counted the multiples of \text{LCM}(3,5)=15 twice, once for 3 and once for 5.

There are \left\lfloor \frac{46}{15}\right\rfloor =3 numbers divisible by 15, so there are 24-3=21 numbers divisible by 3 or 5. \rightarrow \boxed{\text{B}}

See Also

1991 AJHSME (Problems • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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
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