To find the maximum value of
$n$
for which
${40}^n$
divides
$60!$
, we first express the number 40 in terms of its prime factors:

$40=2^3\times 5$

Therefore, we can write:
${40}^n={(2^3\times 5)}^n=2^{3n}\times 5^n$

For
${40}^n$
to divide
$60!$
, the prime factor 2 must appear at least
$3n$
times in the prime factorization of
$60!$
, and the prime factor 5 must appear at least
$n$
times.

We can find the exponent of any prime
$p$
in
$N!$
using Legendre's formula:
$E_p(N!)=\lfloor \frac{N}{p}\rfloor +\lfloor \frac{N}{p^2}\rfloor +\lfloor \frac{N}{p^3}\rfloor +\cdots$
where
$\lfloor x\rfloor$
represents the greatest integer less than or equal to
$x$
.

First, let us calculate the exponent of the prime 5 in
$60!$
:
$E_5(60!)=\lfloor \frac{60}{5}\rfloor +\lfloor \frac{60}{25}\rfloor$
$E_5(60!)=12+2=14$
Since
$5^n$
must divide
$60!$
, we have the inequality:
$n\le 14$

Next, let us calculate the exponent of the prime 2 in
$60!$
:
$E_2(60!)=\lfloor \frac{60}{2}\rfloor +\lfloor \frac{60}{4}\rfloor +\lfloor \frac{60}{8}\rfloor +\lfloor \frac{60}{16}\rfloor +\lfloor \frac{60}{32}\rfloor$
$E_2(60!)=30+15+7+3+1=56$
Since
$2^{3n}$
must divide
$60!$
, we must have:
$3n\le 56$
$n\le \lfloor \frac{56}{3}\rfloor =18$

To satisfy both constraints simultaneously, the value of
$n$
must satisfy:
$n\le min(14,18)=14$

Therefore, the maximum value of
$n$
for which
${40}^n$
divides
$60!$
is equal to 14.