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

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GMAT Instructor
Joined: 04 Jul 2006
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Kudos [?]: 158 [0], given: 0

n is a positive integer. Is (k-5)^2>0? (1) n! ends in [#permalink]  21 Nov 2006, 04:18
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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
Director
Joined: 02 Mar 2006
Posts: 582
Location: France
Followers: 1

Kudos [?]: 35 [0], given: 0

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
VP
Joined: 25 Jun 2006
Posts: 1175
Followers: 2

Kudos [?]: 72 [0], given: 0

agree with D and the above explanation.
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1

Kudos [?]: 19 [0], given: 0

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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Impossible is nothing

Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 181 [0], given: 2

(1) how can this be sufficient? if K=5 then (k-5)^2=0?

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
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 181 [0], given: 2

Oh dang!..X&Y good work...i just saw your working

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