GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Sep 2018, 22:17

### 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

# For prime number y, y>3. Which of the following could be the remainder

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 49258
For prime number y, y>3. Which of the following could be the remainder  [#permalink]

### Show Tags

24 Oct 2016, 23:22
00:00

Difficulty:

15% (low)

Question Stats:

79% (00:54) correct 21% (01:20) wrong based on 265 sessions

### HideShow timer Statistics

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

_________________
SVP
Joined: 08 Jul 2010
Posts: 2302
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

### Show Tags

25 Oct 2016, 00:10
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

Checking with prime numbers greater than 3

@y=5, Rem (y^3 / 12) = Rem (5^3 / 12) = 5 (BINGO)

Answer: Option D
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4020
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)
Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

### Show Tags

25 Oct 2016, 08:15
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

y = { 5 , 7 , 11............ }

Plug in a number and check -

5^3/12 = Remainder 5

Thats in the given option , hit it...

Answer will be (D)
_________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )

Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2835
For prime number y, y>3. Which of the following could be the remainder  [#permalink]

### Show Tags

25 Oct 2016, 16:56
1
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
_________________

Jeffery Miller
Head of GMAT Instruction

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Senior Manager
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

### Show Tags

27 Nov 2016, 09:26
1
1
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
Joined: 27 Aug 2016
Posts: 91
Location: India
Schools: HEC Montreal '21
GMAT 1: 670 Q47 V37
GPA: 3
WE: Engineering (Energy and Utilities)
Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

### Show Tags

19 Apr 2017, 11:41
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: 49258
Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

### Show Tags

19 Apr 2017, 11:51
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
Joined: 11 Feb 2017
Posts: 23
Location: India
Schools: SPJ PGPM"17
GMAT 1: 600 Q48 V25
GPA: 3.57
WE: Engineering (Computer Software)
Re: For prime number y, y>3. Which of the following could be the remainder  [#permalink]

### Show Tags

23 May 2017, 13:55
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
Re: For prime number y, y>3. Which of the following could be the remainder &nbs [#permalink] 23 May 2017, 13:55
Display posts from previous: Sort by

# For prime number y, y>3. Which of the following could be the remainder

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.