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Wrong answer ?

by sapuna » Sat Jul 19, 2014 3:25 am
Question : Does the integer k have a factor p such that 1 < p < k ?

1) k > 4!

2) 13! + 2 < or = k < or = 13 ! + 13

1) k > 24 = > k = 25 ( has a factor p = 5 so that 1 < 5 < k ) but k can also be 31 ( does not have a factor other than 31 and 1 ) so 1) is not sufficient

2) by subtracting 13! I get

2 < or = k < or = 13

k can be 4 but it can also be 11 so its not sufficient either

My answer was : Both statements , even when taken together , are not enough to answer the question

The correct answer that was given was : Statement 2 alone is enough to answer the question but Statement 1 isnt.
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by GMATinsight » Sat Jul 19, 2014 4:31 am
sapuna wrote:Question : Does the integer k have a factor p such that 1 < p < k ?

1) k > 4!

2) 13! + 2 < or = k < or = 13 ! + 13
Question : Does the integer k have a factor p such that 1 < p < k ?

K will always have some factor between 1 and k if k is NOT prime therefore

Question Rephrased: Is k a prime number?

Statement 1) k > 4!

i.e. k>24 therefore it could be Prime number or may not be Prime number

INSUFFICIENT

Statement 2) 13! + 2 < or = k < or = 13 ! + 13


k = 13! + 2 but since 13! is a multiple of all number from 1 till 13
therefore 13!+2 = 2(1x3x4xx5x6x7x8x9x10x11x12x13+1) will be divisible 2

Similarly
k = 13! + 3 = 13!+3 = 3(1x2x4x5x6x7x8x9x10x11x12x13+1) will be divisible 3
k = 13! + 4 = 13!+4 = 4(1x2x3x5x6x7x8x9x10x11x12x13+1) will be divisible 4
k = 13! + 5 = 13!+5 = 5(1x2x3x4x6x7x8x9x10x11x12x13+1) will be divisible 5
............
k = 13! + 13 = 13!+13 = 3(1x2x3x4xx5x6x7x8x9x10x11x12+1) will be divisible 13

i.e. k is NEVER PRIME

SUFFICIENT

Answer: Option B
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by sapuna » Sat Jul 19, 2014 5:07 am
Thank you again all !

Joking , thank you GmaTInsight
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by GMATinsight » Sat Jul 19, 2014 5:54 am
sapuna wrote:Thank you again all !

Joking , thank you GmaTInsight
Fun while studying is all that makes one a WINNER in GMAT...

I hope you enjoy the journey of being a winner.

All the best!!!

:mrgreen: :mrgreen: :mrgreen:
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by Brent@GMATPrepNow » Sat Jul 19, 2014 7:11 am
Does the integer k have a factor p such that 1 < p < k ?

(1) k > 4!
(2) 13! + 2 ≤ k ≤ 13! + 13
Target question: Does the integer k have a factor p such that 1 < p < k ?

This question is a great candidate for rephrasing the target question. (We have a free video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100)

Let's look at a few cases to get a better idea of what the target question is asking.
- Try k = 6. Since 2 is a factor of 6, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 10 Since 5 is a factor of 10, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 16. Since 4 is a factor of 14, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 5. Since 1 and 5 are the ONLY factors of 5, we can see that k does NOT have a factor p such that 1<p<k.
Aha, so if k is a prime number, then it CANNOT satisfy the condition of having a factor p such that 1 < p < k
In other words, the target question is really asking us whether k is a non-prime integer (aka a "composite integer")

REPHRASED target question: Is integer k a non-prime integer?

Statement 1: k > 4!
In other words, k > 24
This does not help us determine whether or not k is a non-prime integer? No.
Consider these two conflicting cases:
Case a: k = 25, in which case k is a non-prime integer
Case b: k = 29, in which case k is a prime integer
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 13! + 2 ≤ k ≤ 13! + 13
Let's examine a few possible values for k.

k = 13! + 2
= (13)(12)(11)....(5)(4)(3)(2)(1) + 2
= 2[(13)(12)(11)....(5)(4)(3)(1) + 1]
Since k is a multiple of 2, k is a non-prime integer

k = 13! + 3
= (13)(12)(11)....(5)(4)(3)(2)(1) + 3
= 3[(13)(12)(11)....(5)(4)(2)(1) + 1]
Since k is a multiple of 3, k is a non-prime integer

k = 13! + 4
= (13)(12)(11)....(5)(4)(3)(2)(1) + 4
= 4[(13)(12)(11)....(5)(3)(2)(1) + 1]
Since k is a multiple of 4, k is a non-prime integer

As you can see, this pattern can be repeated all the way up to k = 13! + 13. In EVERY case, k is a non-prime integer

Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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