2017/021) find integer n where $\dfrac{99^n+19^n}{n!}$ is highest

Let us consider $\dfrac{99^x}{x!}$

this value increases for x upto 98 and at 99 the value is same as at 98.

now for the second part $\dfrac{19^x}{x!}$ the value increases for x upto 18 and at x = 19 is same as at x = 18 and then it decreases

for x = 19 to 98 the term $\dfrac{99^x}{x!}$ increases more rapidly as $\dfrac{19^x}{x!}$ decreases
at x = 19 the 1st term is higher by $3.8^{19}$ times and it is much higher
so we need to compare the value at x = 98 and x = 99.
the second term  is fighter at x = 98

so $\dfrac{99^x+19^x}{x!}$ is highest at x = 98