Question 3
The prime factorization of the product of the factors of 15 is $3^2\times5^2$. What is the highest exponent on the prime factorization of the product of the factors of 140? Hint! For any positive integer $n$, the product of all its factors (divisors) is $n^{d(n)/2}$, where $d(n)$ is the number of positive divisors of $n$.