|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
15 Feb 2021, 03:07
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
46% (01:52) correct
54%
(01:55)
wrong
based on 149
sessions
History
Date
Time
Result
Not Attempted Yet
If n is a positive integer, is n(n^2 - 1) is divisible by 24?
(1) n is divisible by 3.
(2) n is odd
(1) n is divisible by 3.
(2) n is odd
XSatishX
Joined: 05 Jul 2020
Last visit: 17 Nov 2022
Posts: 100
Own Kudos:
Given Kudos: 36
Location: India
Concentration: Leadership, General Management
Schools: LBS '24 (D) ISB '23 (II) Tepper '24 (S)
GMAT 1: 650 Q49 V30
GMAT 2: 730 Q48 V41 (Online)
GPA: 3.8
Products:
15 Feb 2021, 03:30
Kudos
Bookmarks
we can write n(n^2-1) as n(n-1)(n+1) as n^2-1 = (n-1)(n+1)
So now we have product of 3 consecutive integers.
As we know, a product of 3 consecutive integers will always be divisible by 3 and 2.
But in order to be divisible by 24, the product should be divisible by 8 and 3.
Stament 1: Not Sufficient
As we know n is divisible by 3 so let us test this condition.
If n = 3 than 2*3*4 = 24 and we get a yes.
if n = 6 than 5*6*7 = 210 and we get a No
Hence this is not sufficient.
Statment 2: Sufficient
n is odd. Let us test this.
If n =1 then 0*1*2 = 0 and 0 is divisible by any integer so yes.
If n = 3 then 2*3*4 is divisible by 24.
If n = 5 then 4*5*6 is divisible by 24.
If n = 7 then 6*7*8 is divisible by 24.
So far we are getting only yes. So now let us look at the logic behind this. If n is odd then both (n-1) and (n+1) will be even. Also in two consecutive even integers, one is always a multiple of 4. For eg: 2&4 or 4&6 or 6&8. So we will always have a 2 and a 4 in the product of both these numbers. This makes (n-1)(n+1) divisible by 8 provided n is odd.
And as we know (n-1)n(n+1) is already divisible by 3 hence it will also be divisible by 3*8 or 24
Answer:B
So now we have product of 3 consecutive integers.
As we know, a product of 3 consecutive integers will always be divisible by 3 and 2.
But in order to be divisible by 24, the product should be divisible by 8 and 3.
Stament 1: Not Sufficient
As we know n is divisible by 3 so let us test this condition.
If n = 3 than 2*3*4 = 24 and we get a yes.
if n = 6 than 5*6*7 = 210 and we get a No
Hence this is not sufficient.
Statment 2: Sufficient
n is odd. Let us test this.
If n =1 then 0*1*2 = 0 and 0 is divisible by any integer so yes.
If n = 3 then 2*3*4 is divisible by 24.
If n = 5 then 4*5*6 is divisible by 24.
If n = 7 then 6*7*8 is divisible by 24.
So far we are getting only yes. So now let us look at the logic behind this. If n is odd then both (n-1) and (n+1) will be even. Also in two consecutive even integers, one is always a multiple of 4. For eg: 2&4 or 4&6 or 6&8. So we will always have a 2 and a 4 in the product of both these numbers. This makes (n-1)(n+1) divisible by 8 provided n is odd.
And as we know (n-1)n(n+1) is already divisible by 3 hence it will also be divisible by 3*8 or 24
Answer:B
27 Feb 2021, 16:13
Kudos
Bookmarks
i went for C though i knew it was a question of n as odd or even? my reasoning was if i take 1 as a value of n then the expression becomes 0. My understanding was that 0 is not a multiple of any value.
please someone clear my doubts.
please someone clear my doubts.
