<s>[INST] Positive divisors of n = $2^{14} \cdot 3^9 \cdot 5^8 \cdot 7^{10} \cdot 11^3 \cdot 13^5 \cdot 37^{10}$ How do I find positive divisors of n that are perfect cubes that are multiples of 2^10 * 3^9 * 5^2 * 7^5 * 11^2 * 13^2 * 37^2
The answer is (1)(1)(2)(2)(1)(1)(3) = 12
I don't understand though because I would have done something like:
2: [(14-10)/3]+1 = 2 (taking the floor)
3: [(9-9)/3]+1 = 1
5: [(8-2)/3]+1 = 3
7: [(10-5)/3]+1 = 2
11: [(3-2)/3]+1 = 1
13: [(5-2)/3]+1 = 2 
37: [(10-2)/3]+1 =3
2*1*3*2*1*2*3  
 [/INST] Since we are required to be a multiple of $2^{10}\cdot3^9\cdot5^2\cdot7^5\cdot11^2\cdot 13^2\cdot37^2$ and also a perfect cube, we know that whatever our divisor is, it must be divisible by $2^{12}\cdot3^9\cdot5^3\cdot7^6\cdot11^3\cdot 13^3\cdot37^3$. Now when you run your counting argument, the permissible ranges of exponents should be small enough to match the provided answer.
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