1995 AHSME Problems/Problem 29

Problem

For how many three-element sets of positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$ ?

$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$

Solution

$2310 = 2\cdot 3\cdot 5\cdot 7\cdot 11$ . The number of ordered triples $(x,y,z)$ with $xyz = 2310$ is therefore $3^5$ , since each prime dividing 2310 divides exactly one of $x,y,z$ .

Three of these triples have two of $x,y,z$ equal (namely when one is 2310 and the other two are 1). So there are $3^5 - 3$ with $x,y,z$ distinct.

The number of sets of distinct integers $\{ a,b,c\}$ such that $abc = 2310$ is therefore $\frac {3^5 - 3}{6}$ (accounting for rearrangement), or $\boxed{40}$ .

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1995 AHSME Problems/Problem 29

Problem

Solution

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