What is the smallest positive integer n such that 2n is a perfect square and 3n is a perfect cube and 5n is a perfect fifth power?

Guest May 15, 2022

#2**0 **

Consider the prime factorization of n. Suppose that \(n = 2^x 3^y 5^z\). (It does not have any other prime factors because n is the smallest one)

Then \(2n = 2^{x + 1} 3^y 5^z\), \(3n = 2^{x} 3^{y + 1} 5^z\), and \(5n = 2^x 3^y 5^{z + 1}\).

2n being a perfect square means all exponents are divisible by 2, i.e., x is odd, y and z are even.

3n being a perfect cube means all exponents are divisible by 3, i.e., x, y + 1, and z are divisible by 3.

5n being a perfect fifth power means all exponents are divisible by 5, i.e., x, y, and z + 1 are divisible by 5.

The smallest odd value of x that is divisible by 3 and 5 is 15.

The smallest even value of y such that y + 1 is divisible by 3 and y is divisible by 5 is 20.

The smallest even value of z such that z is divisible by 3 and z + 1 is divisible by 5 is 24.

Therefore, \(n = 2^{15}3^{20}5^{24}\) is the required answer.

MaxWong May 15, 2022