How many consecutive zeroes are there at the end of 100! (100 factorial)?

Method #1

100! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 9 x 10 â€¦ = 9332621544394415268199238856266700490715968264381621468559296389521759

99932299156089414639761565182862518286253697920827223758251185209168400

0000000000000000000000

Â

Â

Method #2

The general solution in finding the number of zeros in 100!, 1000!, 10 000!, 100 000!, 1 000 000! â€¦

Where,Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  N is one of the following: 100, 1 000, 10 000, â€¦

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Condition: Only accept the term that is when the numerator is greater than the denominator.

Example: Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  where N = 100

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  = Â (Numerator is greater than the denominator.)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Â = Â (Do not accept this, because the numerator is less than the denominator.)

Going back to the equation:

Note: Do not include the term and the terms after that since the denominator is greater than the numerator.

Therefore, there are 24 consecutive zeroes are there at the end of 100!Â

MORE STUDYING TIPS HERE!