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Prime or Composite?

Expert replies
by tanvis1120 » Fri Aug 08, 2014 8:04 pm
Does integer k have a factor p such that 1<p<k?
a. k>4!
b.13!+1<k<13!+14

My answer is B.
I do not know the official answer.
@Expert, right or wrong?
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Source: — Data Sufficiency |

by Brent@GMATPrepNow » Fri Aug 08, 2014 8:20 pm
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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by GMATinsight » Sat Aug 09, 2014 6:16 am
tanvis1120 wrote:Does integer k have a factor p such that 1<p<k?
a. k>4!
b.13!+1<k<13!+14

My answer is B.
I do not know the official answer.
@Expert, right or wrong?
You are going great Tanvi!!!

Your answer is CORRECT... :) ;)
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by tanvis1120 » Sat Aug 09, 2014 6:46 am
Brent@GMATPrepNow wrote:
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
Hi Brent,

I just tested one case for option 2 which is:
13!+13 = 13(12!+1) So, K certainly is non prime as it can be expressed as a product of two numbers other than 1 and itself.
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by Brent@GMATPrepNow » Sat Aug 09, 2014 7:59 am
tanvis1120 wrote: Hi Brent,

I just tested one case for option 2 which is:
13!+13 = 13(12!+1) So, K certainly is non prime as it can be expressed as a product of two numbers other than 1 and itself.
Hi tanvis1120,

If you tested only ONE possible value of k (and found it to be composite), you cannot be certain that ALL possible values of k will be composite.

Here's an analogous question:

Target question: Is n prime?
(1) 3 < n < 7

We can't test only one value here.
If we test the first possible value of n (n = 3), we find it to be prime.
Can I then conclude that statement 1 is sufficient? No. We need to test all of the possible values of n.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by tanvis1120 » Sun Aug 10, 2014 8:09 am
Thanks Brent.. I got the flaw in my approach..
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