BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Wrong answer ?

Expert replies
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.
Join the discussion
Source: — Data Sufficiency |

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
"GMATinsight"Bhoopendra Singh & Sushma Jha
Most Comprehensive and Affordable Video Course 2000+ CONCEPT Videos and Video Solutions
Whatsapp/Mobile: +91-9999687183 l [email protected]
Contact for One-on-One FREE ONLINE DEMO Class Call/e-mail
Most Efficient and affordable One-On-One Private tutoring fee - US$40-50 per hour
Join the discussion

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

Joking , thank you GmaTInsight
Join the discussion

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:
"GMATinsight"Bhoopendra Singh & Sushma Jha
Most Comprehensive and Affordable Video Course 2000+ CONCEPT Videos and Video Solutions
Whatsapp/Mobile: +91-9999687183 l [email protected]
Contact for One-on-One FREE ONLINE DEMO Class Call/e-mail
Most Efficient and affordable One-On-One Private tutoring fee - US$40-50 per hour
Join the discussion

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
Image
Join the discussion