Most of the test dont have many abtract problems like EST sample tests and I think it is good to get used to with calculation
I'm working on it so if you are interested in, let's discuss

Sorry the answer is 5 - there is five numbers satisfying the requirement .* 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
By that logic, if 2400 was one of the answer choices, you would choose that?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