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1995 AHSME Problems/Problem 29

Revision as of 10:18, 8 August 2020 by Skyguy88 (talk | contribs)

Problem

For how many three-element sets of distinct 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 $\frac{3^4 - 1}{2} = 40 \Rightarrow \boxed{C}$

Solution 3

The prime factorization of $2310$ is $2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11.$ Therefore, we have the equation \[abc = 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11,\]where $a, b, c$ must be distinct positive integers and order does not matter. There are $3$ ways to assign each prime number on the right-hand side to one of the variables $a, b, c,$ which gives $3^5 = 243$ solutions for $(a, b, c).$ However, three of these solutions have two $1$s and one $2310,$ which contradicts the fact that $a, b, c$ must be distinct. Because each prime factor appears only once, all other solutions have $a, b, c$ distinct. Correcting for this, we get $243 - 3 = 240$ ordered triples $(a, b, c)$ where $a, b, c$ are all distinct.

Finally, since order does not matter, we must divide by $3!,$ the number of ways to order $a, b, c.$ This gives the final answer, \[\frac{240}{3!} = \frac{240}{6} = \boxed{40}.\]

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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