GRE Prep 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.
If n^m leaves a remainder of 1 after division by 7 for all positive in
[#permalink]
13 Jul 2021, 04:40
13 Jul 2021, 04:40
1
Expert Reply
4
Bookmarks
00:00
Question Stats:
47% (02:49) correct
52% (02:12) wrong based on 17 sessions
Hide Show timer Statistics
If n^m leaves a remainder of 1 after division by 7 for all positive integers n that are not multiples of 7, then m could be equal to :
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If n^m leaves a remainder of 1 after division by 7 for all positive in
[#permalink]
13 Jul 2021, 05:01
13 Jul 2021, 05:01
1
1
Bookmarks
Carcass wrote:
If n^m leaves a remainder of 1 after division by 7 for all positive integers n that are not multiples of 7, then m could be equal to :
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
The most important piece of information is here: "for all positive integers n that are not multiples of 7"
Since 2 is not a multiple of 7, then it must be the case that, for a particular value of m, 2^m leaves a remainder of 1 after division by 7
Let's check the answer choices....
(A) if m = 2, we get 2^2 = 4.
When we divide 4 by 7, we get a remainder of 4. We need a remainder of 1. ELIMINATE A
(B) if m = 3, we get 2^3 = 8.
When we divide 8 by 7, we get a remainder of 1. KEEP B
(C) if m = 4, we get 2^4 = 16.
When we divide 16 by 7, we get a remainder of 2. We need a remainder of 1. ELIMINATE C
(D) if m = 5, we get 2^5 = 32.
When we divide 32 by 7, we get a remainder of 4. We need a remainder of 1. ELIMINATE D
(E) if m = 6, we get 2^6 = 64.
When we divide 64 by 7, we get a remainder of 1. KEEP E
So, the correct answer is either B or E
Now try a different value of n.
How about n = 3
Check the remaining answer choices....
(B) if m = 3, we get 3^3 = 27.
When we divide 27 by 7, we get a remainder of 6. ELIMINATE B
Answer: E
RELATED VIDEO
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1111
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
If n^m leaves a remainder of 1 after division by 7 for all positive in
[#permalink]
02 Oct 2022, 09:08
02 Oct 2022, 09:08
Given that n^m leaves a remainder of 1 after division by 7 for all positive integers n that are not multiples of 7 and we need to find m could be equal to
Let's take each option choice and check which one satisfies the statement
(A) 2
=> \(n^2\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^2\) = 4 divided by 7 gives 4 remainder ≠ 1 => NOT POSSIBLE
(B) 3
=> \(n^3\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^3\) = 8 divided by 7 gives 1 remainder
Let's take n=3 we get
\(3^3\) = 27 divided by 7 gives 6 remainder ≠ 1 => NOT POSSIBLE
(C) 4
=> \(n^4\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^4\) = 16 divided by 7 gives 2 remainder ≠ 1 => NOT POSSIBLE
(D) 5
=> \(n^5\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^5\) = 32 divided by 7 gives 4 remainder ≠ 1 => NOT POSSIBLE
(E) 6
So, Answer will be E but let's check for 1 number
=> \(n^6\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^6\) = 64 divided by 7 gives 1 remainder => POSSIBLE
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Remainders
Let's take each option choice and check which one satisfies the statement
(A) 2
=> \(n^2\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^2\) = 4 divided by 7 gives 4 remainder ≠ 1 => NOT POSSIBLE
(B) 3
=> \(n^3\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^3\) = 8 divided by 7 gives 1 remainder
Let's take n=3 we get
\(3^3\) = 27 divided by 7 gives 6 remainder ≠ 1 => NOT POSSIBLE
(C) 4
=> \(n^4\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^4\) = 16 divided by 7 gives 2 remainder ≠ 1 => NOT POSSIBLE
(D) 5
=> \(n^5\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^5\) = 32 divided by 7 gives 4 remainder ≠ 1 => NOT POSSIBLE
(E) 6
So, Answer will be E but let's check for 1 number
=> \(n^6\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^6\) = 64 divided by 7 gives 1 remainder => POSSIBLE
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Remainders
