Given a positive integer 4 , what is the minimum positive integer k such that k! is a multiple of the square of 4!?

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The answer is 8 because 4!*4! = 24*24 = 576 and 8!=40320 , and it is divisible by 576

40320/576=70

Furthermore, 8 is the answer because 7!=5040 is not divisible by 576.

I got the answer by trial and error, I noticed that the numbers may get too big if it would involve larger factorials, is there a formula for this that also minimize the values, or......can it be minimized?

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You can consider the prime factorizations of the factorials. $4!$ has three factors of $2$, one from $2$ and two from $4$, so $4!^2$ has six factors. $7!$ has only four, one each from $2,6$ and two from $4$, so it doesn't work. The general case is given by Legendre's formula.