does the integer k have a factor p such that 1<p<k ?
1. k>4!
2. 13!+2<k<13!+13
gmat prep inequality
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- rommysingh
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Target question: Does the integer k have a factor p such that 1 < p < k ?Does the integer k have a factor p such that 1 < p < k ?
(1) k > 4!
(2) 13! + 2 ≤ k ≤ 13! + 13
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
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.
does the integer k have a factor p such that 1<p<k ?
1. k>4!
2. 13!+2<k<13!+13
Transforming the original condition and the question, we need to find whether k has p as a factor or not and whether k=not prime number and k=composite number?�
In case of 1), the answer is yes for k=25 and the answer is no for k=29. Therefore the condition is not sufficient.
In case of 2), k=13!+3=3(13*12*......*4*2*1+1) yes
k=13!+4=4(13*12*......*5*3*2*1+1) yes
k=13!+5=5(13*12*......*6*4*...*1+1) yes
k=13!+6=6(13*12*......*7*5*...*1+1) yes
k=13!+7=7(13*12*......*8*6*...*1+1) yes
k=13!+8=8(13*12*......*9*7*...*1+1) yes
k=13!+9=9(13*12*...*10*8*...*1+1) yes
k=13!+10=10(13*12*11*9*...*1+1) yes
k=13!+11=11(13*12*10*...*1+1) yes
k=13!+12=12(13*11*...*1+1) yes
The answers are all yes, therefore the condition is sufficient and the answer is B. Transforming the original condition and the question, following variable approach method 1, practically solves 30% of DS questions.
Math Revolution : Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World's First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
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See our Youtube demo (https://www.youtube.com/watch?v=R_Fki3_2vO8)
Remember equal number of variables and independent equations ensures a solution.
does the integer k have a factor p such that 1<p<k ?
1. k>4!
2. 13!+2<k<13!+13
Transforming the original condition and the question, we need to find whether k has p as a factor or not and whether k=not prime number and k=composite number?�
In case of 1), the answer is yes for k=25 and the answer is no for k=29. Therefore the condition is not sufficient.
In case of 2), k=13!+3=3(13*12*......*4*2*1+1) yes
k=13!+4=4(13*12*......*5*3*2*1+1) yes
k=13!+5=5(13*12*......*6*4*...*1+1) yes
k=13!+6=6(13*12*......*7*5*...*1+1) yes
k=13!+7=7(13*12*......*8*6*...*1+1) yes
k=13!+8=8(13*12*......*9*7*...*1+1) yes
k=13!+9=9(13*12*...*10*8*...*1+1) yes
k=13!+10=10(13*12*11*9*...*1+1) yes
k=13!+11=11(13*12*10*...*1+1) yes
k=13!+12=12(13*11*...*1+1) yes
The answers are all yes, therefore the condition is sufficient and the answer is B. Transforming the original condition and the question, following variable approach method 1, practically solves 30% of DS questions.
Math Revolution : Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World's First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Unlimited Access to over 120 free video lessons - try it yourself (https://www.mathrevolution.com/gmat/lesson)
See our Youtube demo (https://www.youtube.com/watch?v=R_Fki3_2vO8)