# Finding the maximum $k$ such that $(7!)!$ is divisible by $(7!)^{k!}\cdot(6!)!$

divisibilityfactorialproblem solving

If $$(7!)!$$ is divisible by $$(7!)^{k!}\cdot(6!)!$$, then what is the maximum value of $$k$$?

At first glance I couldnt think of anything except Legendre's formula for calculating powers of a prime in a factorial. It would indeed be cumbersome task of calculating all primes that occur in $$(6!)!$$. So I came up with another possible solution but I don't know if it gives me the maximum value of $$k$$ or not. Do let me know what you think.

If we see this situation as dividing $$7!$$ different objects in $$6!$$ groups and each group having $$7$$ objects each then the number of ways of making these different groups will be $$\dfrac{(7!)!}{((7!)^{6!})((6!)!)}$$ Now, this will give me $$k=6$$, and this ensures that quotient is a natural number, too.

You've already shown that $$(7!)^{6!}((6!)!)$$ divides $$(7!)!$$. If $$(7!)^{7!}((6!)!)$$ divides $$(7!)!$$, then $$(7!)^{7!}$$ divides $$(7!)!$$, so $$(7!)^{7!}\leq (7!)!$$. However, $$n^n>n!$$ for all $$n>1$$.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}