Sum of exactly n perfect square divisors I'm doing a Number Theory question, and if someone could offer a hint, that would be greatly appreciated;
The question is: Find the sum of the perfect square divisors of the smallest integer with exactly 6 perfect square divisors.
My reasonings: 


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*My method so far has been simply listing out pairs of possible exponents that are multiples of 2, and listing them out, or using combinations to calculate the total number of combinations. Obviously, this brute-force method is not working out well, and if anyone could provide a hint, that would be greatly appreciated.

*I know how to find the product of all divisors, but not their sum. Furthermore, how would one "target" only perfect square divisors? Perhaps complementary counting?


Conversation in the comments is correct, but I would suggest the following approach. First we find the lowest number with exactly $6$ divisors, and then we square that number. This squared number will then have exactly $6$ square divisors.
To find the number, we use the number of divisors formula
$$d\left(\prod p_i^{a_i}\right) = \prod(a_i +1)$$
Now the divisors of $6$ are $6,1$ and $2,3$, so now we choose the lowest primes and attach these exponents to them as follows.
$$2^5=32 \quad 2^2*3=12$$
Now we have the lowest number with exactly $6$ divisors, $12$. So the lowest number with 6 square divisors is $144$. The square divisors are the squares of the divisors of $12$, so now we add those up to get our answer.
$$1^2+2^2+3^2+4^2+6^2+12^2=210$$
