Last visit was: 19 Nov 2025, 07:49 It is currently 19 Nov 2025, 07:49
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
avatar
kunalbh19
avatar
Joined: 27 Nov 2011
Last visit: 20 Jun 2012
Posts: 5
Own Kudos:
110
 [96]
Given Kudos: 4
Location: India
Concentration: Technology, Marketing
GMAT 1: 660 Q47 V34
GMAT 2: 710 Q47 V41
WE:Consulting (Consulting)
GMAT 2: 710 Q47 V41
Posts: 5
Kudos: 110
 [96]
If n is a positive integer, is n^2 - 1 divisible by 24? 25 May 2012, 02:00
9
Kudos
Add Kudos
86
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

56% (02:02) correct 44% (02:10) wrong based on 954 sessions

History

Date
Time
Result
Not Attempted Yet
If n is a positive integer, is n^2 - 1 divisible by 24?

(1) n is a prime number.
(2) n is greater than 191
ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,255
 [86]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,255
 [86]
If n is a positive integer, is n^2 - 1 divisible by 24? 25 May 2012, 02:31
45
Kudos
Add Kudos
40
Bookmarks
Bookmark this Post
If n is a positive integer, is n^2 - 1 divisible by 24?

(1) n is a prime number --> if n=2, then the answer is NO but if n=5, then the answer is YES. Not sufficient.

(2) n is greater than 191. Clearly insufficient (consider n=24^2 for a NO answer and n=17^2 for an YES answer).

(1)+(2) Given that n is a prime number greater than 191 so n is odd and not a multiple of 3. n^2-1=(n-1)(n+1) --> out of three consecutive integers (n-1), n and n+1 one must be divisible by 3, since it's not n then it must be either (n-1) or (n+1), so (n-1)(n+1) is divisible by 3. Next, since n is odd then (n-1) and (n+1) are consecutive even numbers, which means that one of them must be a multiple of 4, so (n-1)(n+1) is divisible by 2*4=8. We have that (n-1)(n+1) is divisible by both 3 and 8 so (n-1)(n+1) is divisible by 3*8=24. Sufficient.

Answer: C.

Hope it's clear.
User avatar
paranoidvik
User avatar
Joined: 17 Jul 2013
Last visit: 18 Jan 2015
Posts: 19
Own Kudos:
33
 [10]
Given Kudos: 183
Location: India
WE:Information Technology (Healthcare/Pharmaceuticals)
Products:
My Reviews
Posts: 19
Kudos: 33
 [10]
If n is a positive integer, is n^2 - 1 divisible by 24? 19 Jun 2014, 11:20
4
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
This question can be rephrased as follows:

Is n^2 - 1 = 24k ?

Is n^2 = 24k+1? (24 k + 1 is odd)
so question is: Is n^2 = odd? (since n is positive integer, and if n^2 = odd then n must be odd)
so question is : Is n = odd ?

statement 1: n is prime : If n = 2 = even, if n = 3 = odd .Not suff
Statement 2: n > 191 , n = 192 = even, n = 193 (odd) . Not suff

1 + 2 : n is prime and > 191 = n is odd. Sufficient
C
General Discussion
User avatar
321kumarsushant
User avatar
Joined: 01 Nov 2010
Last visit: 19 Jun 2015
Posts: 138
Own Kudos:
172
 [2]
Given Kudos: 44
Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE:Marketing (Manufacturing)
Posts: 138
Kudos: 172
 [2]
If n is a positive integer, is n^2 - 1 divisible by 24? 27 May 2012, 01:59
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
If n is a positive integer, is n^2 - 1 divisible by 24?

(1) n is a prime number --> if n=2, then the answer is NO but if n=5, then the answer is YES. Not sufficient.

(2) n is greater than 191. Clearly insufficient (consider n=24 for a NO answer and n=17 for an YES answer).

(1)+(2) Given that n is a prime number greater than 191 so n is odd and not a multiple of 3. n^2-1=(n-1)(n+1) --> out of three consecutive integers (n-1), n and n+1 one must be divisible by 3, since it's not n then it must be either (n-1) or (n+1), so (n-1)(n+1) is divisible by 3. Next, since n is odd then (n-1) and (n+1) are consecutive even numbers, which means that one of them must be a multiple of 4, so (n-1)(n+1) is divisible by 2*4=8. We have that (n-1)(n+1) is divisible by both 3 and 8 so (n-1)(n+1) is divisible by 3*8=24. Sufficient.

Answer: C.

Hope it's clear.
Show more

bunuel can you please explain the colored part of your answer.
if n is not there, why it n-1 & n+1 must be divisible by 3.
it may be possible that only n is divisible by 3 example; n-1=2,n=3,n+1=4 i.e 2,3,4
if 3 is not there,, why 2x4 must be divisible by 3 ??
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,255
 [2]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,255
 [2]
If n is a positive integer, is n^2 - 1 divisible by 24? 27 May 2012, 04:38
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
321kumarsushant
Bunuel
If n is a positive integer, is n^2 - 1 divisible by 24?

(1) n is a prime number --> if n=2, then the answer is NO but if n=5, then the answer is YES. Not sufficient.

(2) n is greater than 191. Clearly insufficient (consider n=24 for a NO answer and n=17 for an YES answer).

(1)+(2) Given that n is a prime number greater than 191 so n is odd and not a multiple of 3. n^2-1=(n-1)(n+1) --> out of three consecutive integers (n-1), n and n+1 one must be divisible by 3, since it's not n then it must be either (n-1) or (n+1), so (n-1)(n+1) is divisible by 3. Next, since n is odd then (n-1) and (n+1) are consecutive even numbers, which means that one of them must be a multiple of 4, so (n-1)(n+1) is divisible by 2*4=8. We have that (n-1)(n+1) is divisible by both 3 and 8 so (n-1)(n+1) is divisible by 3*8=24. Sufficient.

Answer: C.

Hope it's clear.

bunuel can you please explain the colored part of your answer.
if n is not there, why it n-1 & n+1 must be divisible by 3.
it may be possible that only n is divisible by 3 example; n-1=2,n=3,n+1=4 i.e 2,3,4
if 3 is not there,, why 2x4 must be divisible by 3 ??
Show more

For (1)+(2) we have that n is odd and not a multiple of 3. Next, (n-1), n and n+1 represent three consecutive integers. Out of ANY three consecutive integers one is always divisible by 3, we know that it's not n, so it must be either n-1 or n+1.

Hope it's clear.
User avatar
sandal85
User avatar
Joined: 28 Feb 2012
Last visit: 05 Feb 2014
Posts: 20
Own Kudos:
274
 [3]
Given Kudos: 1
Posts: 20
Kudos: 274
 [3]
If n is a positive integer, is n^2 - 1 divisible by 24? 30 May 2012, 09:25
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I have another solution which came to me.
Clearly first and second statement alone each can not be sufficient to find out whether expression is divisible by 24 or not.
Now considering both, we know that n is prime number which is greater than 191.
All prime numbers except 2 and 3 are in the form of 6n+1 or 6n-1.

n^2 -1 = (6n+1)^2 - 1 or n^2 -1 = (6n - 1)^2 - 1

n^2 -1 = 36n^2 + 12n or n^2 -1 = 36n^2 - 12n

= 12(n^2 + n) or 12(n^2 - n)

Now we know that this expression is divisible by 12 and this will be divisible by 24 if (n^2+n) or (n^2-n) is even numbers.
As we know that n is prime number which is greater than 191 so it has to be odd so
n^2 will also be odd and now we can say that n^2+n will be even (odd+odd = even).
Same way n^2 - n will be even.

So this expression will definitely divisible by 24.

Bunuel - Please let me know if I am doing any wrong over here.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,255
 [5]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,255
 [5]
If n is a positive integer, is n^2 - 1 divisible by 24? 31 May 2012, 02:04
5
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sandal85
I have another solution which came to me.
Clearly first and second statement alone each can not be sufficient to find out whether expression is divisible by 24 or not.
Now considering both, we know that n is prime number which is greater than 191.
All prime numbers except 2 and 3 are in the form of 6n+1 or 6n-1.

n^2 -1 = (6n+1)^2 - 1 or n^2 -1 = (6n - 1)^2 - 1

n^2 -1 = 36n^2 + 12n or n^2 -1 = 36n^2 - 12n

= 12(n^2 + n) or 12(n^2 - n)

Now we know that this expression is divisible by 12 and this will be divisible by 24 if (n^2+n) or (n^2-n) is even numbers.
As we know that n is prime number which is greater than 191 so it has to be odd so
n^2 will also be odd and now we can say that n^2+n will be even (odd+odd = even).
Same way n^2 - n will be even.

So this expression will definitely divisible by 24.

Bunuel - Please let me know if I am doing any wrong over here.
Show more

Your approach is correct, but when expressing prime number \(n\), you shouldn't use the same variable (\(n\)). Also there are some other little mistakes.

It should be: since any prime number greater than 3 could be expressed as \(6k+1\) or\(6k+5\) (\(6k-1\)), where \(k\) is an integer >1, then \(n=6k+1\) or \(n=6k-1\).

\(n^2-1=36k^2+12k=12k(3k+1)\) or \(n^2-1=36k^2-12k=12k(3k-1)\).

We can see that \(n^2-1\) is divisible by 12. Next, notice that if \(k=odd\) then \(3k+1=even\) (\(3k-1=even\)) and if \(k=even\) then \(3k+1=odd\) (\(3k-1=odd\)). So, \(n^2-1\) is also divisible by 2, which means that \(n^2-1\) is divisible by 12*2=24.

Hope it's clear.
User avatar
WoundedTiger
User avatar
Joined: 25 Apr 2012
Last visit: 25 Sep 2024
Posts: 521
Own Kudos:
2,534
 [3]
Given Kudos: 740
Location: India
GPA: 3.21
WE:Business Development (Other)
Products:
My Reviews
Posts: 521
Kudos: 2,534
 [3]
If n is a positive integer, is n^2 - 1 divisible by 24? 26 Jun 2013, 02:30
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kunalbh19
If n is a positive integer, is n^2 - 1 divisible by 24?

(1) n is a prime number.
(2) n is greater than 191
Show more

St 1 : If n =2 then n^2-1 is not divisibly by 24
but if n=7 then n^2-1 is divisible by 24

There is a there property that for any prime (p) greater than 6, p^2 -1 is always divisible by 6
Option A,D ruled out
St 2 says n>191 again if n is not prime than n^2-1 may or may not be divisible by 24 but if n is prime and greater than 191 then it surely divisible by 24
since 2 choices are possible so option B is ruled out

Combining we get n is prime and n>191 and hence the expression n^2-1 will be divisible by 24 for any value of n

Ans is C
User avatar
mau5
User avatar
Verbal Forum Moderator
Joined: 10 Oct 2012
Last visit: 31 Dec 2024
Posts: 479
Own Kudos:
3,340
 [2]
Given Kudos: 141
Posts: 479
Kudos: 3,340
 [2]
If n is a positive integer, is n^2 - 1 divisible by 24? 26 Jun 2013, 03:16
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kunalbh19
If n is a positive integer, is n^2 - 1 divisible by 24?

(1) n is a prime number.
(2) n is greater than 191
Show more

Any prime number greater than 3 can be represented as \(6k\pm1\) but NOT necessarily vice-versa, where k = 1,2,3..etc

I. If n is prime,\(n^2-1 = (6k+1)^2-1\) --> (6k+2)*6k --> 12*k*(3k+1) . Either k is odd and (3k+1) is even OR k is even and (3k+1) is odd. Either ways, with one even factor in the expression, \(n^2-1\) would always have a factor of 24.

II.If n is prime, \(n^2-1 = (6k-1)^2-1\) --> (6k-2)*6k --> 12*k*(3k-1) . Either k is odd and (3k-1) is even OR k is even and (3k-1) is odd. Either ways, with one even factor in the expression, \(n^2-1\) would always have a factor of 24.

Thus, any prime number n>3,\(n^2-1\) is ALWAYS divisible by 24.

F.S 1- We don't know whether n>3 or not. Insufficient.

F.S 2- We don't know whether n is prime or not. For n= 240 we get a YES, for n = 241, we get a NO.Insufficient.

Taking both togehter, we know that n is prime and n>3. Thus, \(n^2-1\) will always be divisible by 24. Sufficient.
C.
avatar
up4gmat
avatar
Joined: 14 Dec 2012
Last visit: 29 Jun 2022
Posts: 59
Own Kudos:
22
 [1]
Given Kudos: 186
Location: United States
Posts: 59
Kudos: 22
 [1]
If n is a positive integer, is n^2 - 1 divisible by 24? 23 Jul 2013, 21:47
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
If n is a positive integer, is n^2 - 1 divisible by 24?

(1) n is a prime number --> if n=2, then the answer is NO but if n=5, then the answer is YES. Not sufficient.

(2) n is greater than 191. Clearly insufficient (consider n=24^2 for a NO answer and n=17^2 for an YES answer).

(1)+(2) Given that n is a prime number greater than 191 so n is odd and not a multiple of 3. n^2-1=(n-1)(n+1) --> out of three consecutive integers (n-1), n and n+1 one must be divisible by 3, since it's not n then it must be either (n-1) or (n+1), so (n-1)(n+1) is divisible by 3. Next, since n is odd then (n-1) and (n+1) are consecutive even numbers, which means that one of them must be a multiple of 4, so (n-1)(n+1) is divisible by 2*4=8. We have that (n-1)(n+1) is divisible by both 3 and 8 so (n-1)(n+1) is divisible by 3*8=24. Sufficient.

Answer: C.

Hope it's clear.
Show more


amazing explanation,Bunuel!You rock!
User avatar
JeffTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 2,977
Own Kudos:
8,389
 [1]
Given Kudos: 1,646
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert
Expert reply
Posts: 2,977
Kudos: 8,389
 [1]
If n is a positive integer, is n^2 - 1 divisible by 24? 27 Nov 2017, 19:16
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kunalbh19
If n is a positive integer, is n^2 - 1 divisible by 24?

(1) n is a prime number.
(2) n is greater than 191
Show more

We need to determine whether (n^2 - 1)/24 = integer. Notice that 24 is 2^3 x 3, or 8 x 3.

Since n^2 - 1 = (n + 1)(n - 1), when n is odd and not a multiple of 3, we will have the product of two consecutive even integers, one of which is a multiple of 3, and thus n^2 - 1 is divisible by 24. For instance, when n is 5, we have 4 x 6, which is divisible by 24.

Statement One Alone:

n is a prime number.

When n is 5, we see that 24/24 = integer; however, when n = 2, 3/24 does not equal an integer. Statement one is not sufficient to answer the question.

Statement Two Alone:

n is greater than 191.

If n = 192, then n^2 - 1 will be odd and will not be divisible by 24. If n = 199, then n^2 - 1 = (199 + 1)(199 - 1) = 200 x 198 is divisible by 24, since 200 is divisible by 8 and 198 is divisible by 3. Statement two is not sufficient to answer the question.

Statements One and Two together:

From both statements, we see that n is a prime that is greater than 191, and thus it satisfies the case that n is odd and not a multiple of 3. So, n^2 - 1 will be divisible by 24.

Answer: C
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,994
 [1]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,994
 [1]
If n is a positive integer, is n^2 - 1 divisible by 24? 26 Oct 2023, 00:54
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kunalbh19
If n is a positive integer, is n^2 - 1 divisible by 24?

(1) n is a prime number.
(2) n is greater than 191
Show more

First, look at this post: https://anaprep.com/arithmetic-factors- ... -integers/

\((n^2 - 1) = (n-1)(n+1)\)

(1) n is a prime number.
n could be 2 in which case 2^2 - 1 is not divisible by 24.
n could be 5 in which case 5^2 - 1 is divisible by 24.

(2) n is greater than 191
Clearly insufficient.

Using both, n is a large prime number. It will be odd and will have no factors other than 1 and itself.
So (n-1) and (n+1) would be even and one of them would have to be divisible by 4. Also, since n is not divisible by 3, one of (n-1) and (n+1) will have to be.
Hence (n-1) and (n+1) together will have 3 and 8 as factors and hence will be divisible by 24.

Answer (C)


Note that the only relevance of n > 191 is to tell us that n is neither 2 nor 3. For all other prime numbers, n^2 - 1 will be divisible by 24.
Say n = 11;
(n-1) = 10 (has 2 as factor)
(n+1) = 12 (has 4 and 3 as factors)
Hence the product 10 * 12 will be divisible by 24.
User avatar
youcouldsailaway
User avatar
Joined: 08 Nov 2022
Last visit: 07 Feb 2024
Posts: 63
Own Kudos:
52
Given Kudos: 57
Location: India
Concentration: Finance, Entrepreneurship
Schools: Wharton '26 Cornell '26 Ross '26 Fuqua '27 HBS '26 Stern '26 Marshall '26 Tuck '26
GMAT 1: 590 Q37 V34
GMAT 2: 660 Q40 V40
GPA: 3
Schools: Wharton '26 Cornell '26 Ross '26 Fuqua '27 HBS '26 Stern '26 Marshall '26 Tuck '26
GMAT 2: 660 Q40 V40
Posts: 63
Kudos: 52
If n is a positive integer, is n^2 - 1 divisible by 24? 28 Oct 2023, 03:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I arrived at the answer by substituting n for different values and found that for any prime number greater than or equal to 5, the condition is met.

St. 1 says the number is prime but n can be 2 or 3 too, which don't satisfy the condition. Not sufficient.

St. 2 says n is greater than 191, but we can't be sure about the exact value of n. Not sufficient.

C is the answer because after combining both we know N will meet the condition.
User avatar
aronbhati
User avatar
Joined: 13 Jul 2024
Last visit: 18 Nov 2025
Posts: 40
Own Kudos:
1
Given Kudos: 90
Location: India
Concentration: Operations, Finance
GPA: 8.25
Posts: 40
Kudos: 1
If n is a positive integer, is n^2 - 1 divisible by 24? 17 Jan 2025, 22:25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Mam,

in Statement 2: n is greater than 191. I am not able to understand while solving option C, you haven't use n greater than 191
KarishmaB
kunalbh19
If n is a positive integer, is n^2 - 1 divisible by 24?

(1) n is a prime number.
(2) n is greater than 191

First, look at this post: https://anaprep.com/arithmetic-factors- ... -integers/

\((n^2 - 1) = (n-1)(n+1)\)

(1) n is a prime number.
n could be 2 in which case 2^2 - 1 is not divisible by 24.
n could be 5 in which case 5^2 - 1 is divisible by 24.

(2) n is greater than 191
Clearly insufficient.

Using both, n is a large prime number. It will be odd and will have no factors other than 1 and itself.
So (n-1) and (n+1) would be even and one of them would have to be divisible by 4. Also, since n is not divisible by 3, one of (n-1) and (n+1) will have to be.
Hence (n-1) and (n+1) together will have 3 and 8 as factors and hence will be divisible by 24.

Answer (C)


Note that the only relevance of n > 191 is to tell us that n is neither 2 nor 3. For all other prime numbers, n^2 - 1 will be divisible by 24.
Say n = 11;
(n-1) = 10 (has 2 as factor)
(n+1) = 12 (has 4 and 3 as factors)
Hence the product 10 * 12 will be divisible by 24.
Show more
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,994
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,994
If n is a positive integer, is n^2 - 1 divisible by 24? 18 Jan 2025, 01:43
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Note the highlighted below.

aronbhati
Hi Mam,

in Statement 2: n is greater than 191. I am not able to understand while solving option C, you haven't use n greater than 191
KarishmaB
kunalbh19
If n is a positive integer, is n^2 - 1 divisible by 24?

(1) n is a prime number.
(2) n is greater than 191

First, look at this post: https://anaprep.com/arithmetic-factors- ... -integers/

\((n^2 - 1) = (n-1)(n+1)\)

(1) n is a prime number.
n could be 2 in which case 2^2 - 1 is not divisible by 24.
n could be 5 in which case 5^2 - 1 is divisible by 24.

(2) n is greater than 191
Clearly insufficient.

Using both, n is a large prime number. It will be odd and will have no factors other than 1 and itself.
So (n-1) and (n+1) would be even and one of them would have to be divisible by 4. Also, since n is not divisible by 3, one of (n-1) and (n+1) will have to be.
Hence (n-1) and (n+1) together will have 3 and 8 as factors and hence will be divisible by 24.

Answer (C)


Note that the only relevance of n > 191 is to tell us that n is neither 2 nor 3. For all other prime numbers, n^2 - 1 will be divisible by 24.
Say n = 11;
(n-1) = 10 (has 2 as factor)
(n+1) = 12 (has 4 and 3 as factors)
Hence the product 10 * 12 will be divisible by 24.
Show more
Moderators:
Bunuel
Math Expert
105389 posts
kingbucky
496 posts