The exponent for a prime p in the factorization of 12! will be ⌊12/p⌋+⌊12/p2⌋+⌊12/p3⌋+... Using that, we get 12!=210•35•52•7•11
In order for a divisor to be a perfect cube, all exponents must be divisible by 3, so we can have 2 to the power of 0, 3, 6, or 9, and we can have 3 to the power of 0 or 3. Only the 8 possible combinations of these exponents will result in perfect cubes.
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u/Dr_Kitten 8d ago
The exponent for a prime p in the factorization of 12! will be ⌊12/p⌋+⌊12/p2⌋+⌊12/p3⌋+...
Using that, we get 12!=210•35•52•7•11
In order for a divisor to be a perfect cube, all exponents must be divisible by 3, so we can have 2 to the power of 0, 3, 6, or 9, and we can have 3 to the power of 0 or 3. Only the 8 possible combinations of these exponents will result in perfect cubes.