Last visit was: 20 Jul 2024, 09:37 It is currently 20 Jul 2024, 09:37
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
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94430
Own Kudos [?]: 642520 [8]
Given Kudos: 86708
Send PM
GMAT Club Legend
GMAT Club Legend
Joined: 08 Jul 2010
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Posts: 6020
Own Kudos [?]: 13810 [0]
Given Kudos: 125
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
Send PM
Board of Directors
Joined: 11 Jun 2011
Status:QA & VA Forum Moderator
Posts: 6049
Own Kudos [?]: 4767 [0]
Given Kudos: 463
Location: India
GPA: 3.5
WE:Business Development (Commercial Banking)
Send PM
Target Test Prep Representative
Joined: 04 Mar 2011
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Posts: 3036
Own Kudos [?]: 6599 [1]
Given Kudos: 1646
Send PM
For prime number y, y>3. Which of the following could be the remainder [#permalink]
1
Kudos
Expert Reply
Bunuel wrote:
For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12?

A. 2
B. 3
C. 4
D. 5
E. 6


Let’s test the first prime number greater than 3, which is 5.

5^3 = 125

125/12 = 10 remainder 5

Answer: D
Senior Manager
Senior Manager
Joined: 13 Oct 2016
Posts: 299
Own Kudos [?]: 781 [2]
Given Kudos: 40
GPA: 3.98
Send PM
Re: For prime number y, y>3. Which of the following could be the remainder [#permalink]
1
Kudos
1
Bookmarks
Bunuel wrote:
For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12?

A. 2
B. 3
C. 4
D. 5
E. 6


Prime number greater than 3 can be expressed as \(6k + 1\) or \(6k -1\).

\(\frac{(6k ± 1)^3}{12} = \frac{6^3k^3 ± 3*6^2k^2 + 3*6k ± 1}{12} = \frac{0 + 0 + 18k ± 1}{12} = \frac{6k ± 1}{12}\)

Let’s check first case \(6k + 1\)

k=1 (7/12) rem=7
k=2 (13/12) rem =1
k=3 (19/12) rem = 7
k =4 (25/12) rem = 1

we have cycle of remainders \(7\) and \(1\).

Next \(6k – 1\)

k=1 (5/12) rem=5
k=2 (11/12) rem =11
k=3 (17/12) rem = 5
k =4 (23/12) rem = 11

again cycle of only two remainders \(5\) and \(11\).

The only remainders after division by 12 cube of prime number \(y > 3\) can have are \(1\), \(5\), \(7\) and \(11\).

Checking our options.

Answer D.
Manager
Manager
Joined: 27 Aug 2016
Posts: 74
Own Kudos [?]: 17 [0]
Given Kudos: 149
Location: India
GMAT 1: 670 Q47 V37
GPA: 3
WE:Engineering (Energy and Utilities)
Send PM
Re: For prime number y, y>3. Which of the following could be the remainder [#permalink]
vitaliyGMAT wrote:
Bunuel wrote:
For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12?

A. 2
B. 3
C. 4
D. 5
E. 6


Prime number greater than 3 can be expressed as \(6k + 1\) or \(6k -1\).

\(\frac{(6k ± 1)^3}{12} = \frac{6^3k^3 ± 3*6^2k^2 + 3*6k ± 1}{12} = \frac{0 + 0 + 18k ± 1}{12} = \frac{6k ± 1}{12}\)

Let’s check first case \(6k + 1\)

k=1 (7/12) rem=7
k=2 (13/12) rem =1
k=3 (19/12) rem = 7
k =4 (25/12) rem = 1

we have cycle of remainders \(7\) and \(1\).

Next \(6k – 1\)

k=1 (5/12) rem=5
k=2 (11/12) rem =11
k=3 (17/12) rem = 5
k =4 (23/12) rem = 11

again cycle of only two remainders \(5\) and \(11\).

The only remainders after division by 12 cube of prime number \(y > 3\) can have are \(1\), \(5\), \(7\) and \(11\).

Checking our options.

Answer D.


I have a doubt here please....prime numbers graeter than 3 can be expressed as 6k+1..putting k=4 we get 25, which is not a prime num,ber....am I missing something here?
Math Expert
Joined: 02 Sep 2009
Posts: 94430
Own Kudos [?]: 642520 [0]
Given Kudos: 86708
Send PM
Re: For prime number y, y>3. Which of the following could be the remainder [#permalink]
Expert Reply
saurabhsavant wrote:
vitaliyGMAT wrote:
Bunuel wrote:
For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12?

A. 2
B. 3
C. 4
D. 5
E. 6


Prime number greater than 3 can be expressed as \(6k + 1\) or \(6k -1\).

\(\frac{(6k ± 1)^3}{12} = \frac{6^3k^3 ± 3*6^2k^2 + 3*6k ± 1}{12} = \frac{0 + 0 + 18k ± 1}{12} = \frac{6k ± 1}{12}\)

Let’s check first case \(6k + 1\)

k=1 (7/12) rem=7
k=2 (13/12) rem =1
k=3 (19/12) rem = 7
k =4 (25/12) rem = 1

we have cycle of remainders \(7\) and \(1\).

Next \(6k – 1\)

k=1 (5/12) rem=5
k=2 (11/12) rem =11
k=3 (17/12) rem = 5
k =4 (23/12) rem = 11

again cycle of only two remainders \(5\) and \(11\).

The only remainders after division by 12 cube of prime number \(y > 3\) can have are \(1\), \(5\), \(7\) and \(11\).

Checking our options.

Answer D.


I have a doubt here please....prime numbers graeter than 3 can be expressed as 6k+1..putting k=4 we get 25, which is not a prime num,ber....am I missing something here?


Any prime number \(p>3\) when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).

So any prime number \(p>3\) could be expressed as \(p=6n+1\) or\(p=6n+5\) or \(p=6n-1\), where n is an integer >1.

But:
Not all number which yield a remainder of 1 or 5 upon division by 6 are prime, so vise-versa of above property is not correct. For example 25 yields a remainder of 1 upon division be 6 and it's not a prime number.

Hope it's clear.
Intern
Intern
Joined: 11 Feb 2017
Posts: 18
Own Kudos [?]: 7 [0]
Given Kudos: 29
Location: India
Schools: SPJ PGPM"17
GMAT 1: 600 Q48 V25
GPA: 3.57
WE:Engineering (Computer Software)
Send PM
Re: For prime number y, y>3. Which of the following could be the remainder [#permalink]
prime number 5 , 5^3 /12 = 25*5 /12 => 1*5/12 ( since 25/12 will give rem as 1) => 5 is rem

Option D is answer
Intern
Intern
Joined: 13 Jun 2020
Posts: 23
Own Kudos [?]: 0 [0]
Given Kudos: 95
Location: India
Send PM
Re: For prime number y, y>3. Which of the following could be the remainder [#permalink]
We can check it by plugging prime numbers greater than 3.

For Ex. Take prime number 5. Y^3 = 125 and divide by 12 then remainder become 5. Hence, Answer is D

Hope it helps !
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 34039
Own Kudos [?]: 853 [0]
Given Kudos: 0
Send PM
Re: For prime number y, y>3. Which of the following could be the remainder [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: For prime number y, y>3. Which of the following could be the remainder [#permalink]
Moderator:
Math Expert
94430 posts