1991 AJHSME Problems/Problem 9

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 $5$ . $\rightarrow \boxed{\text{B}}$

1991 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1991 AJHSME Problems/Problem 9

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