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.
Answers
MathMate
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
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
unowen
Thank you.
MathMate
you're welcome!