Here's the question you clicked on:
geerky42
How many numbers in the form \(a^4\), where \(a \in \mathbb{Z}^+\) divide \(3! \times 4! \times 7!\) ?
i don't think there are too many since you need primes to the power of 4
included in \(3!4!7!\) is \(2^7\) and \(3^4\) all other primes are to lower powers
i am not certain but on the basis of prime factorization i only see \[2^4,3^4,(2\times 3)^4\]
oops i miscounted!! it is \(2^8\) and \(3^4\)
so maybe there are 4 all together, \[2^4, 3^4, (2^2)^4,(2\times 3)^4, (2^2\times 3)^4\]
well that is actually 5, not 4