* 99 zeros will happen exactly when there are 99 10's ==> 99 5's and 99 2's (we can ignore 10, coz k! is a multiple of primes)
* since number 5 is more than 2, then when 5's reach 99, the 2's will be more than 99 .. eg : 5! = 1.2.3.2.2.5 ==> there are 3 2's when 5's is 1.
* so we can focus only on the 5's.
* For example x! is the number when the first 99 5's happens. then before the 5's reach 100, there can only be number (x+1)!, (x+2)!, (x+3)!, (x+4)!.. that (x+5)! will make 5's reach 100..
* so, there should be only x,x+1,x+2,x+3,x+4 numbers when 5's reach 99
It took me about an hour to solve this problem BUT if we choose p=7>5 then 7^4-1=2401-1=2400 so the answer must be 5) 240 -------> easy and save a lot of time
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