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Hard tricky question

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Hard tricky question [#permalink]  20 Jul 2010, 20:13
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If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
6
12
14
42
56

Clearly there are 12 multiples of 5, therefore, 12 zeros. I don't understand....
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Re: Hard tricky question [#permalink]  20 Jul 2010, 20:23
knabi wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
6
12
14
42
56

Clearly there are 12 multiples of 5, therefore, 12 zeros. I don't understand....

\frac{60}{5}+\frac{60}{5^2}=12+2=14.

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Re: Hard tricky question [#permalink]  20 Jul 2010, 20:25
this formula is very useful...
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Re: Hard tricky question [#permalink]  21 Jul 2010, 00:34
Bunuel, can you elaborate more on this formula..
I would have found the number of times 5 is repeated to get to the answer..
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Re: Hard tricky question [#permalink]  21 Jul 2010, 01:52
Hi Bunuel,

The explanation on the link is great..Is there a similar explanation on remainders and power cycle concepts?

Thanks,
SS
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Re: Hard tricky question [#permalink]  21 Jul 2010, 10:00
SS2010 wrote:
Hi Bunuel,

The explanation on the link is great..Is there a similar explanation on remainders and power cycle concepts?

Thanks,
SS

Check Math: Number Theory for powers and cycles: math-number-theory-88376.html

Check sriharimurthy's post on remainders: compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html

temp33 wrote:
Bunuel, can you elaborate more on this formula..
I would have found the number of times 5 is repeated to get to the answer..

Hope it helps.
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Re: Hard tricky question   [#permalink] 21 Jul 2010, 10:00
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