Question

How many numbers in the form \(a^4\), where \(a \in \mathbb{Z}^+\) divide \(3! \times 4! \times 7!\) ?

Answer 1

i don't think there are too many since you need primes to the power of 4

Answer 2

included in \(3!4!7!\) is \(2^7\) and \(3^4\) all other primes are to lower powers

Answer 3

i am not certain but on the basis of prime factorization i only see
\[2^4,3^4,(2\times 3)^4\]

Answer 4

oops i miscounted!!
it is \(2^8\) and \(3^4\)

Answer 5

so maybe there are 4 all together,
\[2^4, 3^4, (2^2)^4,(2\times 3)^4, (2^2\times 3)^4\]

Answer 6

well that is actually 5, not 4