It is currently 22 Jun 2017, 05:27

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1262
n is a positive integer. Is (k-5)^2>0? (1) n! ends in [#permalink]

### Show Tags

21 Nov 2006, 05:18
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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: 575
Location: France

### Show Tags

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
VP
Joined: 25 Jun 2006
Posts: 1166

### Show Tags

21 Nov 2006, 08:48
agree with D and the above explanation.
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK

### Show Tags

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...
_________________

Impossible is nothing

Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

### Show Tags

25 Nov 2006, 11:41
(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: 3357
Location: New York City
Schools: Wharton'11 HBS'12

### Show Tags

25 Nov 2006, 11:44
Oh dang!..X&Y good work...i just saw your working

I am soo out of touch ....with GMAT..
25 Nov 2006, 11:44
Display posts from previous: Sort by