1981 AHSME Problems/Problem 7

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 $1$ and $100$ , so the answer is $\textbf{(B)}\ 1.$

-edited by coolmath34

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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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1981 AHSME Problems/Problem 7

Problem 7

Solution

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