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Does the integer k have a factor p such that 1<p<k?

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Does the integer k have a factor p such that 1<p<k?

by eloka » Wed Jun 02, 2010 6:12 pm
1) k>4!
2) 13! + 2 <= k <= 13! + 13

Please help me =(

1) I understand the statement one is not unique because 4! = 4*3*2*2*1 = 24
It means k> 24..so K can be any number and from there we cant determine the p.

2) I am confused.

13! + 2 = we know that 2 must be a factor of 13! because 2 is inside 13!
13! + 13 = we know that 13 must be a factor of 13! because 13 is inside 13!

However, how can we determine P????

Lets make it like this..

1<p<k

k= 13!+2, 13! +3, 13!+4,..................., 13!+13

so for P we can have 2, 3, 4,5,6,7,8,9,10,11,12,13. The answer is more than 1. I still dont get it!!

*with tears and blood*

at least 450 gmat score, I will be happy!!

Warm wishes,

eloka
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Source: — Data Sufficiency |
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by Patrick_GMATFix » Wed Jun 02, 2010 7:24 pm
Where is the Q from please?
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by eloka » Wed Jun 02, 2010 7:28 pm
Does the integer k have a factor p such that 1<p<k?

1) k>4!
2) 13! + 2 <= k <= 13! + 13

Please help me =(

1) I understand the statement one is not unique because 4! = 4*3*2*2*1 = 24
It means k> 24..so K can be any number and from there we cant determine the p.

2) I am confused.

13! + 2 = we know that 2 must be a factor of 13! because 2 is inside 13!
13! + 13 = we know that 13 must be a factor of 13! because 13 is inside 13!

However, how can we determine P????

Lets make it like this..

1<p<k

k= 13!+2, 13! +3, 13!+4,..................., 13!+13

so for P we can have 2, 3, 4,5,6,7,8,9,10,11,12,13. The answer is more than 1. I still dont get it!!

*with tears and blood*

at least 450 gmat score, I will be happy!!

Warm wishes,

eloka
lavachoco.blocked
Warm wishes to all of you who are striving to conquer the GMAT

Eloka
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https://lavachoco.blocked
All I need is 450 GMAT score!! thats all, Geez
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by eloka » Wed Jun 02, 2010 7:31 pm
and its from OG 11 number 155 Data sufficiency
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by Patrick_GMATFix » Wed Jun 02, 2010 7:39 pm
Thanks eloka,

For anyone looking for the Q, it's #153 in OG11. Detailed Solution & Take-Away is attached

-Patrick
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by eloka » Wed Jun 02, 2010 7:55 pm
gracias!
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by Patrick_GMATFix » Wed Jun 02, 2010 10:06 pm
you're welcome.

Read your blog entry. Keep your head up! I have just one question for you: do YOU want to go to B-school? Because if you don't want to go yourself, you might not like it very much.

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by eloka » Thu Jun 03, 2010 3:04 am
I look forward to pursue an MBA/Master of Science program with a concentration in Marketing.

I really want to get accepted in a decent business school. I know my capability and I realize its impossible to reach a high GMAT score. Since I am a normal human being who also wants to pursue an MBA program with my ,unfortunately, limited brain capacity and low IQ below normal people standard, I have researched for some schools that willing to accept potential students with "certain criteria".

I also talked to several MBA advisors and they told me if I can achieve 450 at least, they may consider my application. I am also aware...that reach 450 wont guarantee I will get accepted, but I do believe GMAT is not about everything.

With that being said, I have one more extra month to study for the GMAT and we will see how it goes. I will try my best, however, if it didnt work out...I will surrender and I will look for AACBSP accredited school who does not require GMAT.
Warm wishes to all of you who are striving to conquer the GMAT

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by vinay89 » Sat Jun 05, 2010 3:26 pm
@patrick

Hey patrick...I had made it till figuring out I needed to look for a non prime number...

But with statement 2 say

13!+2 is 2n+2...it is divisible by 2 yes....but how do I figure out if it is a prime number or not?
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by mathai » Tue Jun 08, 2010 7:05 am
Hey man,

i guess the answer to that is B.

Statement 1, alone is not sufficient, since for a prime greater than 24, it would not have a factor such that 1<p<k

Statement 2 alone is sufficient, since if we split it out, the values for K are

K = 13!+2, 13! +3, ... 13! + 13
= (1 x 2 x 3 x ... 13) + 2, ( 1 x 2 x 3 x... 13)+3, ... (1 x 2 x ..13) + 13
= 2((1x3x...13)+1), 3((1x2x4x..13)+1), ... 13((1 x 2 x... 12)+1)

therefore, all the factors are between 2 and 13, and less than whatever value of K.

Hope this helps!

Mathai
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by Patrick_GMATFix » Tue Jun 08, 2010 7:59 pm
vinay89 wrote:@patrick

Hey patrick...I had made it till figuring out I needed to look for a non prime number...

But with statement 2 say

13!+2 is 2n+2...it is divisible by 2 yes....but how do I figure out if it is a prime number or not?
Remember that a prime number is a number that can only be divided by 1 and itself. If a number is divisible by 2, then it's not prime (unless it's the number 2 itself). Therefore, if you determine that 2 is a factor of your number, and if your number does not equal 2, then your number is non-prime.

Does that answer your question?
-Patrick
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