Jul 20 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes Jul 26 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56300

If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
[#permalink]
Show Tags
26 Apr 2016, 01:02
Question Stats:
55% (02:32) correct 45% (02:26) wrong based on 150 sessions
HideShow timer Statistics
If x = n + 5, y = n + 3, and x, y, and n are positive integers, what is the remainder when xy is divided by 6? (1) When n is divided by 6, the remainder is 3. (2) When n is divided by 3, there is no remainder.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Manager
Joined: 03 May 2015
Posts: 183
Location: South Africa
Concentration: International Business, Organizational Behavior
GPA: 3.49
WE: Web Development (Insurance)

Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
[#permalink]
Show Tags
26 Apr 2016, 03:26
Bunuel wrote: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what is the remainder when xy is divided by 6?
(1) When n is divided by 6, the remainder is 3. (2) When n is divided by 3, there is no remainder. xy = n^2 + 8n + 15 A) When n is divided by 6, the remainder is 3. From what's stated, n is an odd multiple of 3. n^2 = odd multiple of 3 8n = even multiple of 3= multiple of 6 15 = odd multiple of 3 so adding them together, even multiple of 3 = multiple of 6 So either A or D now coming to B) n^2 + 8n + 15 if n's a muliple of 3, when it's odd, n's an even multiple of 3. But when n's even it's an odd multiple of 3. So can't confirm if it's a multiple of 6 Answer : A
_________________
Kudos if I helped



Retired Moderator
Joined: 13 Apr 2015
Posts: 1676
Location: India
Concentration: Strategy, General Management
GPA: 4
WE: Analyst (Retail)

Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
[#permalink]
Show Tags
26 Apr 2016, 03:53
Given: x = n + 5 y = n + 3 Remainder when xy/6 = ?
St1: n = 6k + 3 xy = (6k + 8)*(6k + 6) > 6(6k + 8)(k + 1) > Divisible by 6 Sufficient
St2: n = 3k xy = (3k + 5)*(3k + 3) = 3*(3k + 5)(k + 1) > Divisible by 6 when k is odd and not divisible by 6 when k is even Not sufficient
Answer: A



Current Student
Status: It`s Just a pirates life !
Joined: 21 Mar 2014
Posts: 230
Location: India
Concentration: Strategy, Operations
GPA: 4
WE: Consulting (Manufacturing)

Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
[#permalink]
Show Tags
26 Apr 2016, 04:27
Bunuel wrote: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what is the remainder when xy is divided by 6?
(1) When n is divided by 6, the remainder is 3. (2) When n is divided by 3, there is no remainder. A : satisfies divisiblity by 6 for n=3,9,15...all B: satisfies divisibility for n=3,6,9 but not for n=0.
_________________
Aiming for a 3 digit number with 7 as hundredths Digit



Intern
Joined: 21 Oct 2015
Posts: 46

Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
[#permalink]
Show Tags
28 Apr 2016, 22:46
Balajikarthick1990 wrote: Bunuel wrote: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what is the remainder when xy is divided by 6?
(1) When n is divided by 6, the remainder is 3. (2) When n is divided by 3, there is no remainder. A : satisfies divisiblity by 6 for n=3,9,15...all B: satisfies divisibility for n=3,6,9 but not for n=0. Hi, 'N' can not be 0 as per question stem. N must be positive integer and 0 is not a positive integer. Hope this helps.



Intern
Joined: 14 Mar 2014
Posts: 20

If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
[#permalink]
Show Tags
28 Apr 2016, 23:20
xy= (n+3)(n+5) for div by 6 ~~ should be div by 2 &3 statement 1: n is gives remainder =3 when div by 6 . so n+3 is div by 6. therefore xy is div by 6 sufficient
Statement 2: n is div by 3 > if n is odd multiple of 3 then n+3 is div by 6. so xy is div by 6 > id n is even multiple of 3 then (n+3) becomes odd multiple of 3 which is not div by 2 or 6 also since n is even n+5 will be odd which is again not multiple of 2. So InSUfficient
Hence A



IIMA, IIMC School Moderator
Joined: 04 Sep 2016
Posts: 1361
Location: India
WE: Engineering (Other)

If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
[#permalink]
Show Tags
01 Sep 2018, 21:42
Hi chetan2u PKN pushpitkc niks18 VeritasKarishmaDid Vyshak used distributive property of division ie if (a+b) is divisible by c, then a and b are EACH divisible by c, and the fact that a quotient ie k is always an integer.
_________________
It's the journey that brings us happiness not the destination. Feeling stressed, you are not alone!!



VP
Status: Learning stage
Joined: 01 Oct 2017
Posts: 1028
WE: Supply Chain Management (Energy and Utilities)

If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
[#permalink]
Show Tags
01 Sep 2018, 21:57
adkikani wrote: Hi chetan2u PKN pushpitkc niks18 VeritasKarishmaDid Vyshak used distributive property of division ie if (a+b) is divisible by c, then a and b are EACH divisible by c, and the fact that a quotient ie k is always an integer. Hi adkikani, He has factored out from each of the terms in the expression 6k+6. st1: When n is divided by 6, the remainder is 3. This can be written in the form, n=6k+3 So, x=n+5=6k+3+5=6k+8 And y=n+3=6k+3+3= 6k+6=6(k+1)Hence x*y=(6k+8)*6(k+1)= 6(6k+8)*(k+1) (a*b*c=b*a*c=c*a*b) So, xy is a multiple of 6 or, xy is divisible by 6. sufficient. st2: When n is divided by 3, there is no remainder. or, n=3k+0 So, x=n+5=3k+5 And y=n+3=3k+3=3(k+1) Hence x*y=(3k+5)*3(k+1)=3*(3k+5)(k+1) a) The above expression is divisible by 3. So, xy will be divisible by 6 when either 3k+5 or k+1 is a multiple of 2, otherwise xy isn't divisible by 6. So, insufficient Hope it helps.
_________________
Regards,
PKN
Rise above the storm, you will find the sunshine



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9442
Location: Pune, India

Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
[#permalink]
Show Tags
03 Sep 2018, 05:34
adkikani wrote: Hi chetan2u PKN pushpitkc niks18 VeritasKarishmaDid Vyshak used distributive property of division ie if (a+b) is divisible by c, then a and b are EACH divisible by c, and the fact that a quotient ie k is always an integer. By the way, there is no such property. (3+4) is divisible by 7 but neither 3 is nor 4 is.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



VP
Joined: 09 Mar 2018
Posts: 1002
Location: India

Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
[#permalink]
Show Tags
19 Feb 2019, 01:21
Bunuel wrote: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what is the remainder when xy is divided by 6?
(1) When n is divided by 6, the remainder is 3. (2) When n is divided by 3, there is no remainder. for statement 1 to satisfy we get the cases such as 3,9,15 When substituted in x = n + 5, y = n + 3, gives a product divisible by 6 for statement 2 to satisfy, we get the cases such as 3,6,9,12,15 When substituted in x = n + 5, y = n + 3, gives a product divisible by 6 When substituted in x = n + 5, y = n + 3, does not give a product divisible by 6 A
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.
Quote which i can relate to. Many of life's failures happen with people who do not realize how close they were to success when they gave up.




Re: If x = n + 5, y = n + 3, and x, y, and n are positive integers, what
[#permalink]
19 Feb 2019, 01:21






