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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 : $625$

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3 Answers

up vote 6 down vote accepted

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.

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Suppose $p$ is a prime.If one needs to find the number of factors of $p^k$,all one needs to do is to count the numbers in the set $S=\{p^0=1,p,p^2,\dots p^k\}$.There are $k+1$ elements in all and hence $k+1$ factors of $p^k$.

if you have a number of the form $J=p_{1}^{k_1}p_{2}^{k_2}\dots p_{n}^{n}$, you can clearly see that the number of factors of J is $(k_{1}+1)(k_{2} +1)\dots(k_{n}+1)$.

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I find the best approach for these problems, when confused, is to work with smaller examples. Suppose $a$ and $b$ each have 1 prime factor (ie, they are prime). Then how many factors does $ab$ have? Well, since they are distinct, we know $1$, $a$, $b$, and $ab$ all divide $ab$ evenly. And in fact, nothing more can possible divide $ab$.

So the answer above is $(1+1)(1+1)$ as expected.

Now just increase the number of factors and the number of integers to match your question. If you get lost on the way, post in the comments and we'll help you out!

Addendum: You have to be careful when the number of prime factors (5 in your problem above) is not itself prime: then things don't work out as cleanly. The fact that your problem stipulates that each natural number has 5 prime factors immediately implies each number is $p^4$ for some prime $p$. That's a crucial point.

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