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If the number of trailing zeros in n! is there [#permalink]
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kkannan2 wrote:
Hi chetan2u, can you please explain your reasoning more? How do you conclude that n! has 3 more zeroes than (n-1)! whenever n! is a multiple of 5!, and further, how do you know that 625 must be eliminated, but not the two multiples of 125 that are greater than 625 but still 3 digit integers? Thoroughly confused on how to start this one. I know that the trailing zeroes depend on the number of 5s in an expression, since the number of 2s is always greater, but I'm not sure how to apply that in this case. A more thorough explanation could help me visualize how to solve this one easier.­

­Consider the examples below:

The number of trsiling zeros in 124! is:

124/5 + 124/25 = 24 + 4 = 28.

The number of trsiling zeros in 125! is:

125/5 + 125/25 + 125/125 = 25 + 5 + 1= 31.

The number of trsiing zeros of 125! is three more than the number of trailing zeros of 124!

However:

The number of trsiling zeros in 624! is:

624/5 + 624/25 + 624/125 = 124 + 24 + 4 = 152.

The number of trsiling zeros in 625! is:

625/5 + 625/25 + 625/125 + 625/675 = 125 + 25 + 5 + 1 = 156.

The number of trsiing zeros of 625! is four more than the number of trailing zeros of 624!

 ­
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If the number of trailing zeros in n! is there [#permalink]
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