Question

2x is a perfect square, 3x a perfect cube, and 5x a perfect 5th power. Find the sum of the exponents in the prime factorization of the smallest such positive integer x. Thank you.

Answer 1

2,3,5 are coprime to each other.

Say the prime factorization is
x=(2^p)(3^q)(5^r)
where p>0,q>0,r>0 for x>0

To satisfy "2x is a perfect square" =>
mod(p+1,2)=0
mod(q,3)=0
mod(r,5)=0
=> the smallest solution is therefore p=15

Similarly for "3x is a perfect cube" =>
mod(q,2)=0
mod(q+1,3)=0
mod(q,5)=0
=> the smallest solution is q=20

For "5x is a perfect 5th power",
mod(r,2)=0
mod(r,3)=0
mod(r,5+1)=0
=> the smallest solution is 24
So
x=(2^15)(3^20)(5^24)
=6810125783203125000000000000000
and p+q+r=15+20+24=59

Answer 2

Thank you.

Answer 3

you're welcome!