if the number $K$ has $45$ divisors, and the number $K^2$ has $M$ divisors, what is the sum of all possible values of $M$?

Question

if the number $K$ has $45$ divisors, and the number $K^2$ has $M$ divisors, what is the sum of all possible values of $M$?

My try follows.

$45=15×3$ ; $45 = 45×1$ ; $45 = 9×5$ .

Then $k$ has $3$ possible prime factorizations :

$K= p^{14} ×q^2$ and so $k^2= p^{28}×q^4$ so #of divisors of $K^2 = 29×5=145$ .

In the same manner

$K^2$ would have $89$ or $17×9=153$ divisors

So the sum of all possible values of $M$ is $89+145+153= 387$.

Is my answer right?

If not, please help me understand my fault. Thank you for your help.

Answer 1

You're on the right lines. You've correctly found that $145$, $89$ and $153$ are possible values for the number of divisors of $K^2$. However, there is another possibility since $K^2$ could have three different prime factors.