# How many divisors can $bx$ have, given the number of divisors of $b$ and $x$?

May I ask you for a little help about a problem from number theory:

The numbers $x$ and $b$ have exactly 15 resp. 3 divisors. How many divisors could the numbers i) $7x$, ii) $6x$, iii) $x^{2}$, iv) $bx$ have?

I know that I can use the divisor function $d(n)=\sum\limits_{d\mid n}1$ and I also know that $d(nm)=d(n)d(m)$ if $(n,m)=1$. The problem to solve i) is that I don't know if $7$ is a divisor of $x$. Thank you in advance.

• The idea for e.g. (i) is to separate the cases $7 \mid x$ and $7 \nmid x$. A similar separation of cases can be applied to the others. Jun 18 '13 at 12:30
• Some quick observations. $x$ and $b$ are both perfect squares, $b=p^2$ for some prime $p$. Also your problem reads: "How many divisors COULD the numbers have". I would assume that $7$ is not a divisor of $x$ when solving this problem. Jun 18 '13 at 12:33
• What do you know about $d(nm)$, if $(n,m)$ is $>1$? Do you get an inequality? Jun 18 '13 at 12:52

For 1, if $7 \not | x$, you get a factor $2$ from the $7$. All the existing divisors of $x$ are still divisors of $7x$ and each one multiplied by $7$ is also a divisor.
The general rule is that if $x=p_1^{a_1}p_2^{a_2}\dots p_n^{a_n}$ with the $p_i$ primes, it has $(a_1+1)(a_2+1)\dots (a_n+1)$ factors. So if $7|x$ it will increase the $(a_i+1)$ to $(a_i+2)$ What can be the highest power of $7$ dividing $x$?
I believe you need to consider different possibilities. First think of the different possible prime factorisations for $x$ and $b$. Then consider the cases when certain primes appear in both factors. e.g. for $6x$ you would need to consider the cases when 2 and/or 3 are factors of $x$.