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2005 AMC 10A Problems/Problem 15

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Problem

How many positive cubes divide 3! \cdot 5! \cdot 7! ?

\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6

Solution

3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8...

Therefore, a perfect cube that divides 3! \cdot 5! \cdot 7! must be in the form 2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d} where a, b, c, and d are nonnegative multiples of 3 that are less than or equal to 8, 4, 2 and 1, respectively.

So:

a\in\{0,3,6\} (3 posibilities)

b\in\{0,3\} (2 posibilities)

c\in\{0\} (1 posibility)

d\in\{0\}(1 posibility)


So the number of perfect cubes that divide 3! \cdot 5! \cdot 7! is 3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}

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