1995 AHSME Problems/Problem 29

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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 1

$2310 = 2\cdot 3\cdot 5\cdot 7\cdot 11$ . 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 $\boxed{40}$ .

Solution 2

$2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ . We wish to figure out the number of ways to distribute these prime factors amongst 3 different integers, without over counting triples which are simply permutations of one another.

We can account for permutations by assuming WLOG that $a$ contains the prime factor 2. Thus, there are $3^4$ ways to position the other 4 prime numbers. Note that, with the exception of when all of the prime factors belong to $a$ , we have over counted each case twice, as for when we put certain prime factors into $b$ and the rest into $c$ , we count the exact same case when we put those prime factors which were in $b$ into $c$ .

Thus, our total number of cases is $\frac{3^4 - 1}{2} = 40 \Rightarrow \boxed{C}$

1995 AHSME (Problems)
Preceded by Problem 28	Followed by Problem 30
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1995 AHSME Problems/Problem 29

From AoPSWiki

Contents

Problem

Solution 1

Solution 2

See also