1981 AHSME Problems/Problem 7

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Problem 7

How many of the first one hundred positive integers are divisible by all of the numbers $2$, $3$, $4$, and $5$?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution

The least common multiple of 2, 3, 4, and 5 is $2^2 \cdot 3 \cdot 5 = 60.$ There is only one multiple of 60 between 0 and 100, so the answer is $\textbf{(B)}\ 1.$

-edited by coolmath34