How many consecutive zeroes are there at the end of 100! (100 factorial)?
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 … = 9332621544394415268199238856266700490715968264381621468559296389521759
99932299156089414639761565182862518286253697920827223758251185209168400
0000000000000000000000
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Method #2
The general solution in finding the number of zeros in 100!, 1000!, 10 000!, 100 000!, 1 000 000! …
Where,                N is one of the following: 100, 1 000, 10 000, …
               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
               =  (Numerator is greater than the denominator.)
                =  (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
