If $(100!)base10 is converted into base-b, the number of trailing zeroes is 24.
Find the maximum and minimum value of base b
In order to find the minimum and maximum value of b, in the solution they have found the highest power of 2 ,3, 4, 5 and 8 in 100!.
Also, the minimum value of b turned out to be $5$
The maximum value of b turned out to be $40$ How?
The power with which $p$ occurs in $100!$ is $$v_p:=\left\lfloor \frac{100}p\right\rfloor +\left\lfloor \frac{100}{p^2}\right\rfloor+\left\lfloor \frac{100}{p^3}\right\rfloor+\left\lfloor \frac{100}{p^4}\right\rfloor+\ldots$$ Thus if $b$ is a multiple of $p$, we need $v_p\ge 24$. More generally, if $b$ is a multiple of $p^k$, we need $v_p\ge 24 k$. Additionally, we know that for at least one prime, this inequality must be sharp (or else there would be more than $24$ zeroes), i.e., for some $p$, we have $v_p=24k$ where $p^k\| b$. The smallest valid choice is therefore $x=0$, $y=0$, $z=1$ (i.e., $b=5$), and the largest is $x=4$, $y=2$, $z=1$ (i.e., $b=720$).