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.** 🙂