Last visit was: 23 Apr 2026, 16:55 It is currently 23 Apr 2026, 16:55
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Loser94
Joined: 14 Jan 2018
Last visit: 02 Mar 2023
Posts: 135
Own Kudos:
Given Kudos: 77
Location: India
Concentration: General Management, Entrepreneurship
GMAT 1: 680 Q48 V34
GPA: 3.8
WE:Analyst (Consulting)
GMAT 1: 680 Q48 V34
Posts: 135
Kudos: 172
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
yashikaaggarwal
User avatar
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 29 Mar 2026
Posts: 3,089
Own Kudos:
Given Kudos: 1,510
Location: India
GPA: 4
WE:Analyst (Internet and New Media)
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
TestPrepUnlimited
Joined: 17 Sep 2014
Last visit: 30 Jun 2022
Posts: 1,223
Own Kudos:
1,138
 [2]
Given Kudos: 6
Location: United States
GMAT 1: 780 Q51 V45
GRE 1: Q170 V167
Expert
Expert reply
GMAT 1: 780 Q51 V45
GRE 1: Q170 V167
Posts: 1,223
Kudos: 1,138
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
TheUltimateWinner
What is the value of prime number \(p\)?
1) \(p=n^2-1\) where \(n\) is a positive integer
2) \(p\), \(p+2\), and \(p+4\) are all prime

Statement 1:

\(p = n^2 - 1 = (n - 1)*(n + 1)\) so p cannot be prime unless (n - 1) = 1. In that case we have n = 2 and p = 3. Then p = 3 is the only possible prime input. Sufficient.

Statement 2:

If we look at the divide by 3 world, these values are consecutive in that world with p, p + 2 and p + 4 -> p + 1. Since we have 3 consecutive values, one of them must be divisible by 3 and cannot be prime. Then we cannot have any large p's where p, p+2 and p+4 are all prime. However we have a special case of 3, 5, 7 so p = 3 is the only possible scenario. Sufficient.

Ans: D
User avatar
Mo2men
Joined: 26 Mar 2013
Last visit: 09 May 2023
Posts: 2,426
Own Kudos:
Given Kudos: 641
Concentration: Operations, Strategy
Schools: Erasmus (II)
Products:
Schools: Erasmus (II)
Posts: 2,426
Kudos: 1,508
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Loser94
What is the value of prime number p ?

State 1
p=n^2-1=(n-1)*(n+1) so for every value of n except for 2 this will be a even number means it will have more than 1 factors.
so putting n=2 gives p=3
Sufficient

2) p, p+2, p+4 are prime this can happen if p=3 and@ p=11 both prime number holds the equation so not sufficient .

IMO-A


When p=11 then p+4=11+4=15 which is NOT prime number.
User avatar
Gauravji21
Joined: 27 Aug 2020
Last visit: 25 Jan 2024
Posts: 82
Own Kudos:
Given Kudos: 21
Posts: 82
Kudos: 46
Kudos
Add Kudos
Bookmarks
Bookmark this Post
TestPrepUnlimited
TheUltimateWinner
What is the value of prime number \(p\)?
1) \(p=n^2-1\) where \(n\) is a positive integer
2) \(p\), \(p+2\), and \(p+4\) are all prime

Statement 1:

\(p = n^2 - 1 = (n - 1)*(n + 1)\) so p cannot be prime unless (n - 1) = 1. In that case we have n = 2 and p = 3. Then p = 3 is the only possible prime input. Sufficient.

Statement 2:

If we look at the divide by 3 world, these values are consecutive in that world with p, p + 2 and p + 4 -> p + 1. Since we have 3 consecutive values, one of them must be divisible by 3 and cannot be prime. Then we cannot have any large p's where p, p+2 and p+4 are all prime. However we have a special case of 3, 5, 7 so p = 3 is the only possible scenario. Sufficient.

Ans: D

Posted from my mobile device

Hi..
I understand that if we have three consecutive numbers then one of them will be divisible by 3. But in the questions we have P, P+2, P+4. how come these three numbers can be considered consecutive numbers.
User avatar
rajatchopra1994
Joined: 16 Feb 2015
Last visit: 22 Jun 2024
Posts: 1,052
Own Kudos:
Given Kudos: 30
Location: United States
Posts: 1,052
Kudos: 1,307
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Loser94
What is the value of prime number p ?

State 1
p=n^2-1=(n-1)*(n+1) so for every value of n except for 2 this will be a even number means it will have more than 1 factors.
so putting n=2 gives p=3
Sufficient

2) p, p+2, p+4 are prime this can happen if p=3 and p=11 both prime number holds the equation so not sufficient .

IMO-A

In statement 2
As you quoted When P=11,
P+4 will be 15 which is not a prime number.
The only value which satisfies the condition when P=3
avatar
Tonkotsu
Joined: 10 Oct 2020
Last visit: 31 Mar 2023
Posts: 43
Own Kudos:
Given Kudos: 204
Location: United States (TN)
Concentration: Strategy, Finance
GPA: 3.74
WE:Analyst (Consumer Packaged Goods)
Posts: 43
Kudos: 25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Gauravji21
TestPrepUnlimited
TheUltimateWinner
What is the value of prime number \(p\)?
1) \(p=n^2-1\) where \(n\) is a positive integer
2) \(p\), \(p+2\), and \(p+4\) are all prime

Statement 1:

\(p = n^2 - 1 = (n - 1)*(n + 1)\) so p cannot be prime unless (n - 1) = 1. In that case we have n = 2 and p = 3. Then p = 3 is the only possible prime input. Sufficient.

Statement 2:

If we look at the divide by 3 world, these values are consecutive in that world with p, p + 2 and p + 4 -> p + 1. Since we have 3 consecutive values, one of them must be divisible by 3 and cannot be prime. Then we cannot have any large p's where p, p+2 and p+4 are all prime. However we have a special case of 3, 5, 7 so p = 3 is the only possible scenario. Sufficient.

Ans: D

Posted from my mobile device

Hi..
I understand that if we have three consecutive numbers then one of them will be divisible by 3. But in the questions we have P, P+2, P+4. how come these three numbers can be considered consecutive numbers.



I was confused by this 'consecutive' thing also, but looked up some math properties.

We know already that \(P, P + 2, P + 4 \) is a small list of 3 consecutive odd integers.
If we want to see it another way, can rewrite this equation here \( 2n + 1, 2n + 3, 2n + 5 \) where n is an integer.

This shows us that either the first, second, or third number can be divisible by 3 just by nature of being 3 odd numbers apart.
Any integer n (or p in the earlier equation) you plug in will always result in being divisible by three.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,964
Own Kudos:
Posts: 38,964
Kudos: 1,117
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109785 posts
498 posts
212 posts