2k is a perfect square, 3k is a perfect cube, and 5k is a perfect 5th power. What is the sum of the exponents in the prime factorization of the smallest such positive integer k ?

Darkside
Aug 6, 2018

#1**+2 **

This is what I found:

Answer:

59

Explanation:

The prime factorization of k is :

k=pn11⋅pn22⋅...⋅pnmm

That factorization must contain 2, 3 and 5 as prime numbers :

⇒k=2n1⋅3n2⋅5n3

n1 must be odd, multiple of 5, and multiple of 3⇒15

n2 must be 2 modulo 3, even, and multiple of 5⇒20

n3 must be 4 modulo 5, even, and multiple of 3⇒24

So, 15+20+24=59 is the number asked.

Synth
Aug 6, 2018