If $a,b,c,d\in\mathbb{N}$ be distinct. Each of which has exactly five factors, can we determine the number of factors of the product of $a,b,c,d$?
Edit
This is the solution given the in the back of the book I am reading. Does not make sense to me.
If a, b, c and d have five factors each , they are all the fourth powers of prime numbers. Hence their product will have a total of $(4+1)(4+1)(4+1)(4+1) = 625$ factors.
Ans : $625$
Yes!
If you're familiar with how to find the # of factors of a number, then it becomes a matter of prime factorizing each of $a,b,c,d$. Since $5$ is prime itself, it follows that each of the four integers must be of the form $p_i^4$ for some prime $p$ and $i=1,2,3,4$. Now we are given that $a,b,c,d$ are all distinct; namely, each of the $p_i$'s must be distinct. Thus, $abcd= p_1^4 \cdot p_2 ^4 \cdot p_3 ^ 4 \cdot p_4 ^4$, which by the formula, has $(4+1)(4+1)(4+1)(4+1)=625$ factors.