2013 AMC 10B Problem 9

Question

Three positive integers are each greater than

1,

have a product of

27000 ,

and are pairwise relatively prime. What is their sum?

100

137

156

160

165

Put each prime-power factor wholly into one of the three numbers

Pairwise relatively prime means no prime can appear in two different numbers

Only one number is a multiple of

2,

only one is a multiple of

3,

and only one is a multiple of

5.

Since each positive integer is greater than

1,

each must be a multiple of one of these primes.

Now

27000=2^3\cdot 3^3\cdot 5^3.

Therefore, the numbers must be

2^3,3^3,5^3,

and their sum is

160.

Thus, the correct answer is D.

Problem 9 in other years