Finding the Exponent

Relevant wiki: Factorials Problem Solving - Intermediate

Note that $125 = 5^3$ , so $125^n=5^{3n}$

We must find the largest n such that $5^{3n}|100!$ . For this to be true, there must be $3n$ powers of 5 in the prime factorization of $100!$ . As

$100!=100 \cdot 99 ... = 2^2\cdot 5^2 \cdot 3^2 \cdot 11 ...$

we count numbers $\leq 100$ whose prime factorization contains 5, as those are the only numbers which will contribute a 5 to the prime factorization of $100!$ . We count multiples of 25 twice as they contain $5^2$ in their prime factorization, and therefore contribute two 5's to the prime factorization of $100!$ :

$5, 10, 15, 20, 25, 25, 30, 35, 40, 45, 50, 50, 55, 60, 65, 70, 75, 75, 80, 85, 90, 95, 100, 100$

We have 24 numbers, which means that $100!$ contains $5^{24}$ in its prime factorization. Therefore,

$5^{3n}=5^{24}$

$\boxed n=8$

A zero can be formed by the multiplication of 5 and 2. Since 100! contains more than 2's than 5's. We can find the maximum power of 5 contained in 100! $\implies$

$\lfloor\frac{100}{2}\rfloor$ + $\lfloor\frac{100}{4}\rfloor$ + $\lfloor\frac{100}{8}\rfloor$ + $\lfloor\frac{100}{16}\rfloor$ + $\lfloor\frac{100}{32}\rfloor$ + $\lfloor\frac{100}{64}\rfloor$ = $50 + 25+12+6+3+1=97$

$\lfloor\frac{100}{5}\rfloor$ + $\lfloor\frac{100}{25}\rfloor$ = $20$ + $4$ = $24$

So, there are $24$ zero's at the end of 100! $\implies$ $\frac{24}{3}$ = $8$ , since $125=5^{3}$

Generalizing this:

Q. Find the highest power of $k$ that divides $x$ !

Formula to use:

$f(x) =\displaystyle\sum _{ a=1 }^{ \infty } \left\lfloor \frac { x }{ k^{ a } } \right\rfloor$

Using the formula, we have

$k=\boxed{8}$

There are $4$ multiples of $25$ and $20$ multiples of $5$ in $100!$ . This means we could say there $16$ numbers that are divisible by only $5$ and $4$ numbers divisible by $5^{2}$ , that is $24$ $5$ s in total, or $8$ $125$ s in the factorial of $100$ .

As we know, $\frac{100}{5} = 20$ , and this means that $100!=5^{20} \cdot a$ , also we know that $\frac{100}{25}=4$ so $100!=25^{4} \cdot b$ . But from the second expression there is $5^{4}$ that weren't considered in the first expression, so we know $100!=5^{24} \cdot c = 125^{8} \cdot c$ . So the answer is $8$