Question

If the sum of the number of zero at the end of the factorials of first $n$ natural number is 250, the factorial of how many of these natural number will end with an odd number of zeros?

Answer 1

$b!=b\times (b-1)\times (b-2)\times \dots \times (a+1)\times a!$

So, $b!+a!$

$=[b\times (b-1)\times (b-2)\times \dots \times (a+1)+1]\times a!$

So the number of zeroes in the sum

= number of zeroes in first term + number of zeroes in $a!$

Case A: If the product of numbers from b to $(a+1)$ ends in $9$, then the first term will contribute an additional zero.

Case B: Otherwise the number of zeroes will be the same as in $a!$.

We can easily see that the below combinations will fall in case $a$.

$9!+8!$

$19!+18!$

$29!+28!$

And so on.

Coming to the numbers in the question, they will fall in case B. So the number of zeroes in the sum is the same as the number of zeroes in $100!.$

A zero is formed whenever we have a $5$ and a $2$ as prime factors. There are lot of $2$s available but only limited $5$s, from $1$ to $100$.

How many $5$s?

$1$ each from $5,10,20,\dots ,95,100=20$

$1$ extra $5$ in $25,50,75,100=4$

So, $100!$ will have $24$ zeroes.