The "shortcut"
Let's go back to the initial problem of finding $x$ such that $\frac{100!}{3^x}$ is an integer.
Once you find the number of multiples of $3^1$ from $1$ to $100$, continue dividing that result by $3$ and taking the whole number result until you get to zero.
Continuously divide by $3$ and take the whole number result
$$33...11...3...1...0$$
$$33+11+3+1=48$$
Here's another example. How many powers of $5$ are there in $2000!$?
Continuously divide by $5$ and take the whole number result
$$400...80...16...3...0$$
$$400+80+16+3=499$$
This is called Legendre's theorem. You're not expected to know the specific name of the theorem for the GRE.
Bringing it all together
What if we're to find the largest integer $x$ such that the below is an integer?
$$\frac{100!}{24^x}$$
Notice that this problem combines two concepts covered earlier:
- Splitting $24^x$: $24^x = 8^x \times 3^x$
- Decomposing $8^x$ into powers of primes: $8^x = 2^{3x}$
As a result,
$$\frac{100!}{24^x} = \frac{100!}{2^{3x} 3^x}$$
Now the problem is to determine the limiting factor: is it the 2 or the 3? Using the concepts covered earlier, we find that
- $2^{97}$ divides $100!$
- $3^{48}$ divides $100!$
From this, it may look like the limiting factor is the 3, because $48 < 97$. Recall however that we actually have $2^{3x}$ (because of the $8^x$), so the maximum value of $x$ is instead the integer part of $97 \div 3 = 32$. As a result, the limiting factor is actually $2$, because $32 < 48$. Hence the answer is $32$, with $32$ being the largest power of $24$ that divides the factorial of $100$.