1995 AMC 12 Problem 29

Attempt Problem 29 of the 1995 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1995 AMC 12 solutions, or check the answer key.

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29.

For how many three-element sets of positive integers {a,b,c}\{a,b,c\} is it true that a⋅b⋅c=2310?a\cdot b\cdot c=2310?
3232
3636
4040
4343
4545
Answer: C
Concepts:prime factorizationset partitions
Small hint:
Factor 23102310 into distinct primes and assign each prime to one of the factors
Big hint:
Count separately the cases in which one factor is 11 and in which all three factors exceed 11
Solution:
Since 2310=2⋅3⋅5⋅7⋅11,2310=2\cdot3\cdot5\cdot7\cdot11, each prime belongs to exactly one of the three factors. If all factors exceed 1,1, their unordered prime groups form a partition of five objects into three nonempty blocks, counted by S(5,3)=35−3⋅25+36=25. S(5,3)=\frac{3^5-3\cdot2^5+3}{6}=25. If one factor is 1,1, the primes are partitioned into two nonempty blocks, giving S(5,2)=24−1=15.S(5,2)=2^4-1=15. The factors are distinct because their disjoint prime sets differ. Thus the total is 25+15=40,25+15=40, and the correct answer is C.
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