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n is a positive integer. Is (k-5)^2>0? (1) n! ends in

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kevincan
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n is a positive integer. Is (k-5)^2>0? (1) n! ends in [#permalink] New post   21 Nov 2006, 05:18
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n is a positive integer. Is (k-5)^2>0?

(1) n! ends in exactly k zeros.
(2) There are k prime numbers less than 10n



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Re: n is a positive integer. Is (k-5)^2>0? (1) n! ends in [#permalink] New post   21 Nov 2006, 05:54
D

1. n! finishes with five 0 if contains an integer multiplied by 10^5.
10=2.5 so n! has to contain 5 multiples of 2 and 5 multiples of 5.
so n=1.2.3.4.5.6.7.8.9.10....15....20= 20! contains four '5' and more than five '2'. The next possible integer is 25! which contains six '5'.
So there is no n! finishing with five '0'.

SUFFICIENT

(2) There are k prime numbers less than 10n
if n=1 then 10 contains 4 prime numbers.
if n=2 then 20 contains 8 primes.

So again, there are no n satisfaying the condition.

SUFFICIENT
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Re: n is a positive integer. Is (k-5)^2>0? (1) n! ends in [#permalink] New post   21 Nov 2006, 08:48
agree with D and the above explanation.
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Re: n is a positive integer. Is (k-5)^2>0? (1) n! ends in [#permalink] New post   21 Nov 2006, 09:52
Just to give you guys my breakdown for (1) ...

If n=1, then n!=1 and k =0 --> (0-5)^2 > 0 holds true

The only time (k-5)^2>0 will not be true is if k=5. And k will be = to 5 when we have a multiple of 100,000 or 10^5 (ending in five 0s). 100,000 = (5^5) * (2^5)

But I'm not too sure if (1) is SUFFICIENT. I'm thinking if you get the factorial of a bigger number like 40 or 50 (although i didnt compute for this), you might get a number ending in five zeros. Wasn't patient enough to compute tho... Might be too cumbersome for an actual GMAT question so I would go with D. But I think B is a possibility...
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Re: n is a positive integer. Is (k-5)^2>0? (1) n! ends in [#permalink] New post   25 Nov 2006, 11:41
(1) how can this be sufficient? if K=5 then (k-5)^2=0?

not sure i understand your rational.can you please elaborate


karlfurt wrote:
D

1. n! finishes with five 0 if contains an integer multiplied by 10^5.
10=2.5 so n! has to contain 5 multiples of 2 and 5 multiples of 5.
so n=1.2.3.4.5.6.7.8.9.10....15....20= 20! contains four '5' and more than five '2'. The next possible integer is 25! which contains six '5'.
So there is no n! finishing with five '0'.

SUFFICIENT

(2) There are k prime numbers less than 10n
if n=1 then 10 contains 4 prime numbers.
if n=2 then 20 contains 8 primes.

So again, there are no n satisfaying the condition.

SUFFICIENT
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Re: n is a positive integer. Is (k-5)^2>0? (1) n! ends in [#permalink] New post   25 Nov 2006, 11:44
Oh dang!..X&Y good work...i just saw your working :(

I am soo out of touch ....with GMAT..



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Re: n is a positive integer. Is (k-5)^2>0? (1) n! ends in [#permalink]
25 Nov 2006, 11:44
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