GCD Raised too!

Let's say for a positive integer $x$ and a non-negative integer $c$ and prime $p$ , " $p$ -adic order/valuation of $x$ is $c$ ", or write " $v_p(x)=c$ ", to mean $p^c \mid x$ but $p^{c+1}\not \mid x$ ; that is, $p^c$ is the exact power of $p$ that divides $x$ ; (This is, in fact, an standard terminology).

Now, as $13$ is a prime and $gcd(a,b)=13$ , then $13$ -adic order of both $a$ and $b$ is at least $1$ ; and $13$ -adic order is $1$ for at least one of $a$ and $b$ . Here arise $3$ cases.

Case-I : $v_{13}(a)\geq1$ and $v_{13}(b)=1$

Then $gcd(a^3,b)=13$ [As $gcd(a,b)=13$ ]

Case-II : $v_{13}(a)=1$ and $v_{13}(b)=2$

Now, $v_{13}(a^3)=3$ . Then $gcd(a^3,b)=13^2$ [As $gcd(a,b)=13$ ]

Case-III : $v_{13}(a)=1$ and $v_{13}(b)\geq3$

Again, $v_{13}(a^3)=3$ . Then $gcd(a^3,b)=13^3$ [As $gcd(a,b)=13$ ]

So, $(13+13^2+13^3)=\boxed{2379}$ is the answer.

$gcd (a,b)=13$ implies $a$ has $13$ as a factor and $b$ has too. So let $a=p×13^{x}$ and $b=q×13^{y}$ where $p$ and $q$ are coprime, Let $x=1$ . So $y$ can be any natural number.

Now for $gcd (a^{3},b)$ It can be $13, 13^{2},13^{3}$ and not more as $x=1$ .

Let $x>1$ . So $y=1$ because $gcd(a,b)$ is $13$ . So whatever shall be $x$ , $gcd (a^{3},b)$ will be $13$ .

Thus different values= $13,13^2,13^3$ .

Sum= $\boxed {2379}$