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September 3, 2014

September 3, 2014

Posted by **Jon** on Monday, July 30, 2012 at 11:10am.

- math -
**Steve**, Monday, July 30, 2012 at 11:52amyou need 10 to be a factor 100 times.

That means you need 2^100 and 5^100 to be factors.

If [n] is the greatest integer <= n,

n! has 5 as a factor [n/5]+[n/25]+[n/125]+... times.

So, solve [n/5]+[n/25]+[n/125]+[n/625]+[n/3125] = 1000.

if n=4005,

[4005/5]+[4005/25]+[4005/125]+[4005/625]+[4005/3125]

= 801+160+32+6+1 = 1000

Naturally, 2 occurs as a factor many more times than this, so 4005! is the first factorial to end in 1000 zeros.

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