Find the number of consecutive zeros at the end of the following numbers $100!+200!$ To find the tailing zeros of $100!+200!$, I first found $100!$ To be $24$ zeros and $200!$ To be $49$. When we add, the result should be $73$ but the answer is given as $24$ which I cannot understand. Help me out, thanks.

The number of trailing zeros in the decimal representation of $n!$, the factorial of a non-negative integer $n$, can be determined by the formula $$\frac n5+\frac{n}{5^2}+\frac{n}{5^3}+....+\frac{n}{5^k},\mbox{ where $k$ must be chosen such that }5^{k+1}>n$$

Note that the number of tailing zeros in $100!+200!$ is equal to the number of tailing zero's in the smallest factorial. That is because the number of tailing zeros is _different_ in both summands, making sure that the first non-zero digit in $100!$ meets with a zero digit from $200!$ to create the first non-zero digit in the sum. Here the smallest factorial is $100!$ which you already found the tailing zero's to be $24$.

So, the number of tailing zero's in $100!+200!$ is $24$