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Difference between revisions of "1995 AHSME Problems/Problem 29"

(I had a huge solution that was wrong, so I went with Scorpius's solution.)
 
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Revision as of 11:46, 9 January 2008

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 $\boxed{40}$.

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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
All AHSME Problems and Solutions
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