A number theory problem by Paola Ramírez

$\text{If you inspect the Problem closely , you will see a pattern . Let me show you } \\ \text{from } \lfloor{\sqrt1}\rfloor \text{ to } \lfloor{\sqrt{3}}\rfloor\text{We get 1. } \\ \text{from } \lfloor{\sqrt4}\rfloor \text{ to } \lfloor{\sqrt{8}}\rfloor\text{We get 2. } \\ \text{from } \lfloor{\sqrt9}\rfloor \text{ to } \lfloor{\sqrt{15}}\rfloor\text{We get 3. } \\ \text{We can generalize This particular sequence as }\\ \implies 10\displaystyle\prod_{n=1}^{9}n^{2n+1} \\ \text{To find the answer , we must find the power of 5 because power of 2 is more than 5} \\ \text{But i am still going to calculate it } \\ \text{Power of 5 = }5^{11} \\ \text{Power of 2 [when n=2,4,6,8] = }2^5\times2^{18}\times2^{13}\times2^{51}=2^{87} \\ \implies 10\times(5^{11}\times2^{11}) \\ Ans= \boxed{12}$

this problem can be re-written as $\sqrt{100!}$ .We know the fact that 100! has 24 trailing zeroes and so when you square root 100!, the number of trailing zeroes becomes half of the number of trailing zeroes of 100!
[eg.100 has 2 trailing zeroes whereas $\sqrt{100}$ =10 has 1 trailing zero.]

We are seeking for the number of times 5 will appear on the expression. (The 2's don't matter because clearly, there are more 2's than 5's)

So between $36$ and $25$ there are 12 entries of $5's$ hence, our answer.