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

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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

$deleted$

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, $\frac{240}{3!} = \frac{240}{6} = \boxed{40}.$

@@ Line 6: / Line 6: @@
 ==Solution 1==
-<math>2310 = 2\cdot 3\cdot 5\cdot 7\cdot 11</math>. The number of ordered triples <math>(x,y,z)</math> with <math>xyz = 2310</math> is therefore <math>3^5</math>, since each prime dividing 2310 divides exactly one of <math>x,y,z</math>.
+<math>deleted</math>
-Three of these triples have two of <math>x,y,z</math> equal (namely when one is 2310 and the other two are 1). So there are <math>3^5 - 3</math> with <math>x,y,z</math> distinct.
-The number of sets of distinct integers <math>\{ a,b,c\}</math> such that <math>abc = 2310</math> is therefore <math>\frac {3^5 - 3}{6}</math> (accounting for rearrangement), or <math>\boxed{40}</math>.
 ==Solution 2==

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