Answer / sooraj

24

Answer / cnu

24.
exp: 0 is formed by multiplying 5 with 2. so first we find
how many 5's and 2's are there in 100!.
No of 5's : 100/5=20/5=4/5.
20 +4 =24.
No of 2's : 100/2=50/2=25/2=12/2=6/2=3/2=1/2.
50 +25 +12 +6 +3 +1.

It has 24 5's and 2's.

so the no. of zeros=24.

Answer / mohammed ghouse

24

Answer / sumit

It will be minimum of power of 2 and 5 as product of 2 and 5
makes 10.

Power of 2 in this case 94
Power of 5 in this case 24

So answer is 24

Answer / daya

the product of (even number) and 5 results in "0" of unit's
place.

no. of 5's from 1 to 100 = 100/5 = 20
but in 25, 50, 75 and 100, there is one more 5
25 = 5*5*1
50 = 5*5*2
75 = 5*5*3
100= 5*5*4

four 5's extra

therefore no. 0f F's = 20+4 = 24

Answer / rashmi

The product of the first 100 numbers is
933262154439441000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000.

so the number of zeros of the product of the first 100
numbers is 143.

Answer / praveen

if it is 0*1*2........*100 clearly answer is 1;if 4m
1*2*....100 answer is 24

Answer / deepak

Any Number N you want to find the number of zeros in
product expand it in Standard form say for example
(22*25*20*11)=(2*11)*(5*5)*(2*5*2)*(11)
No of 2's= 3
No of 5's= 3 as 2*5=10 so there are 3 combinations of 2*5
so totaly 3 zeros.
For Fafctorial just use this 100! = [100/5]+[100/25]=20+4=24

Answer / kishor

24

Answer / davinder

http://www.calculatorpro.com/calculator/factorial-calculator/

100! = 93,326,215,443,944,102,188,325,606,108,575,267,240,944,254,854,960,571,509,166,910,400,407,995,064,242,937,148,632,694,030,450,512,898,042,989,296,944,474,898,258,737,204,311,236,641,477,561,877,016,501,813,248

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