2005 AMC 10A Problem 15

Question

How many positive cubes divide

3! \cdot 5! \cdot 7!?

2

3

4

5

6

Write

3! \cdot 5! \cdot 7!

as a product of prime powers

A cube divisor uses each prime to an exponent that is a multiple of

3;

count the allowed exponents for each prime

As a product of primes,

3! \cdot 5! \cdot 7! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7.

A cube divisor uses exponents that are multiples of

3:

the exponent of

2

can be

0,

3,

or

6

(

3

choices), the exponent of

3

can be

0

or

3

(

2

choices), and the exponents of

5

and

7

must be

0.

That gives

3 \cdot 2 \cdot 1 \cdot 1 = 6

cubes.

Thus, the correct answer is E.

Problem 15 in other years