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.
On
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 $x$.