Does the integer k have a factor p such that 1 < p < k ?
(1) k > 4!
(2) 13! + 2 ≤ k ≤ 13! + 3
Number Properties
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rephrasing the statement, is K is a prime no.?ela07mjt wrote:Does the integer k have a factor p such that 1 < p < k ?
(1) k > 4!
(2) 13! + 2 ≤ k ≤ 13! + 3
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
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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Refer to the post here >> https://www.beatthegmat.com/integers-pro ... tml#320994
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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.
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.
Ceilidh Erickson
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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
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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The value in red had been posted incorrectly.ela07mjt wrote:Does the integer k have a factor p such that 1 < p < k ?
(1) k > 4!
(2) 13! + 2 ≤ k ≤ 13! + 13
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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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
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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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.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 ).
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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I was being more varietal in approach.
Thanks Anurag.yes it helped
Smiles
Thanks Anurag.yes it helped
Smiles
Anurag@Gurome wrote: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.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 ).
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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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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