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by hemant_rajput » Tue Feb 26, 2013 2:48 am
ela07mjt wrote:Does the integer k have a factor p such that 1 < p < k ?

(1) k > 4!
(2) 13! + 2 ≤ k ≤ 13! + 3
rephrasing the statement, is K is a prime no.?
statement 1:-

k> 4!, i.e. K is greater than 24.

any no. greater than 24(prime or non-prime) is candidate for value of K. So K can be 53 or 50.

Not sufficient.

statement 2:-

13!+ 2 and 13!+3 both are not prime. as 13! is divisible by 2 and adding any even number to a number divisible by 2 will also be divisible by 2. similarly 13! + 3 is also divisible by 3.

Hence 2 is sufficient.

Answer is B
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by Anurag@Gurome » Tue Feb 26, 2013 4:42 am
Refer to the post here >> https://www.beatthegmat.com/integers-pro ... tml#320994
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by ceilidh.erickson » Tue Feb 26, 2013 2:34 pm
The hardest part about this question is figuring out what the question is really asking! A lot of students rephrase this question as "is k divisible by p?" That's not necessarily wrong, but it doesn't actually help to answer the question, because p is just some factor, and not a specified one.

Whenever you're struggling to rephrase a DS question, here are a few things you can do:

- look for CLUE WORDS. Here, the word "factor" is a clue that we're talking about divisibility. If we want to know if it has a factor - if it's divisible by something - that will depend on prime factors.

- think about the "NO" CASE. When would an integer NOT have a factor between 1 and itself? When it's prime! The question must be asking about primes.

Rephrased question: Is k a non-prime?

(1) k > 4!
We know that 4! = 24, so k > 24. This doesn't tell us anything about primes, as there are an infinite number of primes and non-primes greater than 24. Insufficient.

(2) 13! + 2 ≤ k ≤ 13! + 3

Now we have a finite range for k. These numbers are definitely too big to calculate, so let's think about what factorials tell us about divisibility...
Is 13! prime? Definitely not - it has all the factors from 1 to 13: 13*12*11*10*9....

So, is 13! + 2 prime? No! We know that 13! is divisible by 2 (it has several factors of 2 in it, in fact). When we add 2 to a multiple of 2, we get a multiple of 2.

Is 13! + 13 prime? Again, no. We know that 13! is divisible by 13. So a multiple of 13 + a multiple of 13 = a multiple of 13. That sum will be divisible by 13, so it's not prime.

This will be true for any value in that range. 13! + 7 --> a multiple of 7 + a multiple of 7 = a multiple of 7. This is a range of all NON-PRIMES. Sufficient!

The answer is B.
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by Brent@GMATPrepNow » Wed Feb 27, 2013 8:09 am
I just wanted to point out (and formalize) a rule that ceilidh used when determining the sufficiency of statement 2.
The rule is: If k is divisible by d, then (k+d) is divisible by d.
For example, since 238 is divisible by 7, we know that (238+7) is also divisible by 7.


We can also expand the rule to say:
If k is divisible by d, then (k + any multiple of d) is divisible by d.
For example, since 238 is divisible by 7, we know that 238 + (3)(7) is also divisible by 7.
We also know that 238 + (-5)(7) is divisible by 7.

Cheers,
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by GMATGuruNY » Wed Feb 27, 2013 9:35 am
ela07mjt wrote:Does the integer k have a factor p such that 1 < p < k ?

(1) k > 4!
(2) 13! + 2 ≤ k ≤ 13! + 13
The value in red had been posted incorrectly.
I've amended statement 2 to reflect the problem's original intent.

If k has NO FACTOR greater than 1 and less than k, then the only factors of k are 1 AND K ITSELF, implying that k is PRIME.
Question rephrased: Is k prime?

Statement 1: k > 4!
4! = 4*3*2 = 24.
If k = 29, then k is prime.
If k = 25, then k is not prime.
INSUFFICIENT.

Statement 2: 13! + 2 ≤ k ≤ 13! + 13
Apply a bit of REASON.
The values here are HUGE.
There is no way for us to prove that a huge number is prime.
Thus, every value of k that satisfies statement 2 must be NON-PRIME, since it would be impossible for us to prove that any of these values ARE prime.
SUFFICIENT.

The correct answer is B.

A useful take-away:
If a DS problem asks whether a HUGE NUMBER is prime, the answer almost certainly will be NO, since there is no way for us to prove that a huge number actually IS prime (unless we're told rather directly that it has no factors other than 1 and itself).
Last edited by GMATGuruNY on Mon Apr 22, 2013 2:50 am, edited 5 times in total.
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by naadif » Wed Feb 27, 2013 11:44 am
I am not able to get this.I have put my confusion in CAPS below.

Is not the question wants us to check whether K has a factor P and Given is 1<p<k.

option 1: k > 4!
K can be anything greater than 24...K can be 25,26,27,28,29 and so on till infinity.Lets assume k is 25,So we can't be sure enough whether K has a factor P since P can be anything between 1 and 24 in this case.
So not sufficient.

Option 2:13! + 2 ≤ k ≤ 13! + 13
say K= 13!+3 which equals 6227020800+3 =6227020803
NOW HOW CAN WE BE SURE THAT 6227020803 HAS A FACTOR P (GIVEN:1<P<K),P CAN BE ALSO 29,17 OR ANY OTHER PRIME NUMBER THAT WONT DIVIDE K(IN THIS CASE 6227020803 ).
So is not this also insufficient.

Would appreciate the Reply


Brent@GMATPrepNow wrote:I just wanted to point out (and formalize) a rule that ceilidh used when determining the sufficiency of statement 2.
The rule is: If k is divisible by d, then (k+d) is divisible by d.
For example, since 238 is divisible by 7, we know that (238+7) is also divisible by 7.


We can also expand the rule to say:
If k is divisible by d, then (k + any multiple of d) is divisible by d.
For example, since 238 is divisible by 7, we know that 238 + (3)(7) is also divisible by 7.
We also know that 238 + (-5)(7) is divisible by 7.

Cheers,
Brent
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by Anurag@Gurome » Wed Feb 27, 2013 11:53 am
naadif wrote:I am not able to get this.I have put my confusion in CAPS below.

Is not the question wants us to check whether K has a factor P and Given is 1<p<k
...
Option 2:13! + 2 ≤ k ≤ 13! + 13
say K= 13!+3 which equals 6227020800+3 =6227020803
NOW HOW CAN WE BE SURE THAT 6227020803 HAS A FACTOR P (GIVEN:1<P<K),P CAN BE ALSO 29,17 OR ANY OTHER PRIME NUMBER THAT WONT DIVIDE K(IN THIS CASE 6227020803 ).
The question asked whether k has a factor p such that 1 < p < k NOT whether k has all the factors p such that 1 < p < k. This means if there is at least one such p, the answer will be yes.

Hence, if k = 13! + 3 = 1*2*3*4*...*12*13 + 3 = 3*(1*2*4*...*12*13 + 1) = Multiple of 3
Hence, in this case k has a factor p = 3 such that 1 < 3 < k

Hope that helps.
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by naadif » Wed Feb 27, 2013 12:34 pm
I was being more varietal in approach.
Thanks Anurag.yes it helped
Smiles :)
Anurag@Gurome wrote:
naadif wrote:I am not able to get this.I have put my confusion in CAPS below.

Is not the question wants us to check whether K has a factor P and Given is 1<p<k
...
Option 2:13! + 2 ≤ k ≤ 13! + 13
say K= 13!+3 which equals 6227020800+3 =6227020803
NOW HOW CAN WE BE SURE THAT 6227020803 HAS A FACTOR P (GIVEN:1<P<K),P CAN BE ALSO 29,17 OR ANY OTHER PRIME NUMBER THAT WONT DIVIDE K(IN THIS CASE 6227020803 ).
The question asked whether k has a factor p such that 1 < p < k NOT whether k has all the factors p such that 1 < p < k. This means if there is at least one such p, the answer will be yes.

Hence, if k = 13! + 3 = 1*2*3*4*...*12*13 + 3 = 3*(1*2*4*...*12*13 + 1) = Multiple of 3
Hence, in this case k has a factor p = 3 such that 1 < 3 < k

Hope that helps.
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by sana.noor » Thu Jun 06, 2013 11:26 pm
one can simplify this question as "is K prime?" if it is it wont be having any factor between 1 and k but it is not prime then it will have factors between 1 and k. statement 2 is sufficient.
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