Aug 20 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Aug 20 09:00 PM PDT  10:00 PM PDT Take 20% off the plan of your choice, now through midnight on Tuesday, 8/20 Aug 22 09:00 PM PDT  10:00 PM PDT What you'll gain: Strategies and techniques for approaching featured GMAT topics, and much more. Thursday, August 22nd at 9 PM EDT Aug 24 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Aug 25 09:00 AM PDT  12:00 PM PDT Join a FREE 1day verbal workshop and learn how to ace the Verbal section with the best tips and strategies. Limited for the first 99 registrants. Register today! Aug 25 08:00 PM PDT  11:00 PM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE.
Author 
Message 
TAGS:

Hide Tags

eGMAT Representative
Joined: 04 Jan 2015
Posts: 3019

What is the remainder when n2 + 7 is divided by 8, where n is an odd p
[#permalink]
Show Tags
19 Dec 2018, 11:21
Question Stats:
91% (00:58) correct 9% (01:06) wrong based on 175 sessions
HideShow timer Statistics
What is the remainder when \(n^2 + 7\) is divided by 8, where n is an odd prime number? To read all our articles:Must read articles to reach Q51
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



VP
Joined: 07 Dec 2014
Posts: 1229

Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p
[#permalink]
Show Tags
19 Dec 2018, 12:26
EgmatQuantExpert wrote: What is the remainder when \(n^2 + 7\) is divided by 8, where n is an odd prime number?
A. 0 B. 2 C. 3 D. 4 E. 6 let n=3 3^2+7=16 16/8 gives 0 remainder A



GMAT Club Legend
Joined: 18 Aug 2017
Posts: 4476
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p
[#permalink]
Show Tags
20 Dec 2018, 01:30
upon checking relation for n^2+7 ; at n= 1,3,5,7,11.. we can say that the remainder would always be 0 IMO A .. EgmatQuantExpert wrote: What is the remainder when \(n^2 + 7\) is divided by 8, where n is an odd prime number? To read all our articles:Must read articles to reach Q51
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3019

Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p
[#permalink]
Show Tags
01 Jan 2019, 22:46
Solution Given:We are given that • “n” is an odd prime number To find:We need to find, • The remainder when \(n^2 + 7\) is divided by 8 Approach and Working: • If we observe, all the positive integers can be written in the form 4k or + 4k + 1 or 4k + 2 or 4k + 3, where k is a nonnegative integer.
o Out of these, 4k and 4k + 2 cannot be prime numbers, since they are divisible by 2, for any value of k. • Thus, all odd prime numbers must be of the form 4k + 1 or 4k + 3 So, \(n^2\) will be \((4k + 1)^2\) or \((4k + 3)^2\) • And, \(n^2 + 7\) = \((16k^2 + 8k + 8)\) or \((16k^2 + 24k + 16)\) = \(8(2k^2 + k + 1)\) or \(8(2k^2 + 3k + 2)\) • Thus, both these expressions are divisible by 8, for any value of k Therefore, the remainder is 0 Hence the correct answer is Option A. Answer: A
_________________



Senior Manager
Joined: 15 Feb 2018
Posts: 298

What is the remainder when n2 + 7 is divided by 8, where n is an odd p
[#permalink]
Show Tags
28 Jan 2019, 01:03
The question is simple with substitution. The solution above seems to choose 4 arbitrarily. Why not all positive integers can be expressed 2k, 2k+1 resulting in 4(k^2+k+2)?



Intern
Joined: 17 Aug 2018
Posts: 15

Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p
[#permalink]
Show Tags
04 Mar 2019, 18:57
philipssonicare wrote: The question is simple with substitution. The solution above seems to choose 4 arbitrarily. Why not all positive integers can be expressed 2k, 2k+1 resulting in 4(k^2+k+2)? I concur.. Infact when we take 4k+1 we miss out on 3 which is a possible scenario which is not the case while taking 2k+1..



Intern
Joined: 17 Aug 2018
Posts: 15

Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p
[#permalink]
Show Tags
04 Mar 2019, 19:01
EgmatQuantExpert wrote: Solution Given:We are given that • “n” is an odd prime number To find:We need to find, • The remainder when \(n^2 + 7\) is divided by 8 Approach and Working: • If we observe, all the positive integers can be written in the form 4k or + 4k + 1 or 4k + 2 or 4k + 3, where k is a nonnegative integer.
o Out of these, 4k and 4k + 2 cannot be prime numbers, since they are divisible by 2, for any value of k. • Thus, all odd prime numbers must be of the form 4k + 1 or 4k + 3 So, \(n^2\) will be \((4k + 1)^2\) or \((4k + 3)^2\) • And, \(n^2 + 7\) = \((16k^2 + 8k + 8)\) or \((16k^2 + 24k + 16)\) = \(8(2k^2 + k + 1)\) or \(8(2k^2 + 3k + 2)\) • Thus, both these expressions are divisible by 8, for any value of k Therefore, the remainder is 0 Hence the correct answer is Option A. Answer: AWhen we take 4k+1 aren't we missing out on 3 which is a possible scenario??



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3019

What is the remainder when n2 + 7 is divided by 8, where n is an odd p
[#permalink]
Show Tags
04 Mar 2019, 21:26
Quote: The question is simple with substitution. The solution above seems to choose 4 arbitrarily. Why not all positive integers can be expressed 2k, 2k+1 resulting in 4(k^2+k+2)?
We are free to take 2k and 2k + 1 form also. As you mentioned, \((2k + 1)^2 + 7\) will be equal to \(4k^2 + 4k + 8 = 4(k^2 + k) + 8\) Now, for any value of k, \(k^2 + k\) will be either 0 or an even number, hence \(4(k^2 + k)\) will be always divisible by 8. Quote: • Thus, all odd prime numbers must be of the form 4k + 1 or 4k + 3So, \(n^2\) will be \((4k + 1)^2\) or \((4k + 3)^2\)• And, \(n^2 + 7\) = \((16k^2 + 8k + 8)\) or \((16k^2 + 24k + 16)\) = \(8(2k^2 + k + 1)\) or \(8(2k^2 + 3k + 2)\) • Thus, both these expressions are divisible by 8, for any value of k Therefore, the remainder is 0 Quote: When we take 4k+1 aren't we missing out on 3 which is a possible scenario?? If you look at the solution carefully, you will be able to see that we've considered both cases here: the possibility of 4k + 1 as well as the possibility of 4k + 3. The possible scenario of 3 is covered in the case of 4k + 3. Hope this helps.
_________________



Manager
Joined: 18 Jan 2018
Posts: 50

Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p
[#permalink]
Show Tags
07 Mar 2019, 11:18
n is not equal to 2 because n is an odd prime number, n^2 +7 always gives an even number.
Now using n=3,5,7... always results (n^2 +7) = multiples of 8. Hence the remainder of (n^2 +7)/8 is always 0.




Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p
[#permalink]
07 Mar 2019, 11:18






