Suppose $ n = 2^{14} 3^{9} 5^{8} 7^{10} 11^{3} 13^{5} 37^{10} $ , find the number of positive divisors that are both perfect squares and divisible by $ 2^{2}3^{4}5^{2}11^{2}$.
It is quite simple to find both the perfect squares and the number of divisors divisible by $ 2^{2}3^{4}5^{2}11^{2}$separately. But I am confused on how to combine these two properties into one.
To find the number of perfect perfect squares, I find the largest perfect square that is possible and find the maximum combination of perfect square divisors.
$ 2^{14} 3^{8} 5^{8} 7^{10} 11^{2} 13^{4} 37^{10} $ is the largest perfect square.
The largest perfect square can also be written as $((2^{2})^{7}*(3^{2})^{4}*(5^{2})^{4}*(7^{2})^{5}*(11^{2})^{1}*(13^{2})^{2}*(37^{2})^{5}$.
The amount of combinations of the perfect squares is $ (7+1) * (4+1) * (4+1) * (5+1) * (1+1) * (2+1) * (5+1) = 43200$
To find the number of divisors divisible by $ 2^{2}3^{4}5^{2}11^{2}$, I divide $ 2^{14} 3^{9} 5^{8} 7^{10} 11^{3} 13^{5} 37^{10}$ by $ 2^{2}3^{4}5^{2}11^{2}$ which leaves a remainder and the combinations of the remainder is the number of divisors divisible by $2^{2}3^{4}5^{2}11^{2}$.
$ 2^{14} 3^{9} 5^{8} 7^{10} 11^{3} 13^{5} 37^{10}$ / $ 2^{2}3^{4}5^{2}11^{2}$ = $2^{12} 3^{5} 5^{6} 7^{10} 11^{0} 13^{5} 37^{10} $
The amount of combinations divisible by $ 2^{2}3^{4}5^{2}11^{2}$ is $ (12+1) * (5+1) * (6+1) * (10+1) * (0+1) * (5+1) * (10+1) = 396396$
I have tried to find both the perfect squares and the number of divisors divisible by $ 2^{2}3^{4}5^{2}11^{2}$ combined by first finding the largest perfects square and then dividing it by $ 2^{2}3^{4}5^{2}11^{2}$. With the remainder, I find the number of combinations with it.
$(2^{2})^{7}*(3^{2})^{4}*(5^{2})^{4}*(7^{2})^{5}*(11^{2})^{1}*(13^{2})^{2}*(37^{2})^{5}$ / $ 2^{2}3^{4}5^{2}11^{2}$ = $(2^{2})^{6}*(3^{2})^{2}*(5^{2})^{3}*(7^{2})^{5}*(11^{2})^{0}*(13^{2})^{2}*(37^{2})^{5}$
The amount of combinations is $ (6+1) * (2+1) * (3+1) * (5+1) * (0+1) * (2+1) * (5+1) = 9072$
The number 9072 appears to be incorrect however, as I have tried this method with smaller numbers and always ending up with a result that was off.
Any ideas on how find the number of divisors that are perfect squares and divisible by a number?
Best Answer
Any perfect square $S$ has even multiplicity of any given prime number, so if it divides $n = 2^{14} 3^{9} 5^{8} 7^{10} 11^{3} 13^{5} 37^{10}$ it in fact must divide $n' = 2^{14} 3^8 5^8 7^{10} 11^2 13^4 37^{10}$ (all exponents rounded down to even). Now if the perfect square $d=2^{2}3^{4}5^{2}11^{2}$ divides $S$, then $S/d$ is integer and still a perfect square, and it divides $n'/S=2^{12}3^45^67^{10}13^4 37^{10}=m$. Then $\sqrt{S/n'}$ is any divisor of $\sqrt m=2^63^25^37^513^237^5$. You can count those divisors. You did. $9072$ is the right answer.
If you've encountered a smaller example where this approach does not give the right answer, why not present the details of that case?