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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.

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    $\begingroup$ Obviously, only positive divisors are counted; otherwise, since 45 is odd, there is no possible $K$, and the requested sum is zero. $\endgroup$ – fgrieu May 3 '17 at 12:07
  • $\begingroup$ @fgrieu yes you are right , positive divisors $\endgroup$ – Medo May 3 '17 at 12:10
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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.

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  • $\begingroup$ which is ....?? $\endgroup$ – Medo May 3 '17 at 11:59
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    $\begingroup$ $45=3\times 3\times 5$, so it could be $p^2q^2r^4$. $\endgroup$ – Especially Lime May 3 '17 at 12:00
  • $\begingroup$ yes you are right; so the sum =$45+387 =432$ $\endgroup$ – Medo May 3 '17 at 12:05
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    $\begingroup$ @Medo The fourth one should be 225 not 45 as you are doing $(2\cdot2+1)(2\cdot2+1)(2\cdot4+1)=225$. The total is therefore 612. $\endgroup$ – Ian Miller May 3 '17 at 12:10
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    $\begingroup$ Not sure what you are referring to about evenness. The math is as follows: In the fourth case you have $p^2q^2r^4$ from $(2+1)(2+1)(4+1)=45$ divisors. Once you square it you get $p^4q^4r^8$ so you get $(4+1)(4+1)(8+1)=225$ divisors. $\endgroup$ – Ian Miller May 3 '17 at 15:28

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