There are 2 zeros in 10! as, 2*5 has one zero and 10*1 has another.
Using the same logic, we will get that 20! has 4 zeros.
However, 30! has 7 zeros because 5*5=25.
So, we have to look out for 5^x and multiples of them.
40! has 9 zeros.
50! has 11 + 1 zeros because 50=2*5^2.
60! has 14 zeros.
70! has 16 zeros.
80! has 18+1 zeros as 75=3*5^2.
90! has 21 zeros.
100! has 23+1 zeros =24 as 100=4*5^2.
We can find an algorithm:100/5=20
100/5^2=4
20+4=24; It is the number of zeros in 100!.
So now we have a powerful weapon to solve this problem.
No. of zeros in 1000!:1000/5=200
1000/5^2=40
1000/5^3=8
1000/5^4=1.6(We will consider the whole no. part only)
So no. of zeros in 1000!=249.
Similarly no. of zeros in 1100!=220+44+8+1=273.
1200!=240+48+9+1=298
1150!=230+46+9+1=286
1175!=235+47+9+1=292
1170!=234+46+9+1=290.
We got our answer. It is 1170. 🙂