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If n^m leaves a remainder of 1 after division by 7 for all p

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If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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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
[Reveal] Spoiler: OA

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Re: tough 1 [#permalink]

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New post 19 Jun 2011, 11:16
We need to find m for all positive integers.
Therefore, if we take a case of 2 then \(2^3\) and \(2^6\) will leave remainder of 1 after dividing by 7. we will restrict to 6 and not more than that because it is a largest value in the options and we have to select only one of them.

Then, i took a case of 3 ...In that 3^6 leaves remainder of 1 when divided by 7.

Now, we can conclude that the value of m will be 6 because it common value in both cases and one of the given option has to be correct. Hence m=6
thus E

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Re: tough 1 [#permalink]

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AnkitK 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

What is the underlying concept ?


its E
pick up the numbers

ex n = 2
2^1/7 = 2
2^2/7 = 4
2^3/7 = 1
2^4/7 = 4
2^5/7 = 4
2^6 /7 = 1

so m can be 3 or 6
now take n = 3
3^3/7 = 6
3^6/7 = 1

hence 6.

concept:
we know that the remainder when an interger is divided by 7 are 1,2,3,4,5,6 (n-1 concept)
now suppose m =2
the remainder will also be raised by m
1,4,9,16,25,36
divide this by 7 we dont get 1 consistently

try for 3,4,5,.. once u get 6 u will get all 1s
1,64,729,4096, 15625 , 46656) all these number when divided by 7 leaves a remainder 1.

hope this helps.

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Re: tough 1 [#permalink]

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New post 19 Jun 2011, 11:26
Nice explanation Sudhir.
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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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New post 09 Aug 2013, 05:07
Somehow didn't get this question as per below.

n^m = 7Q + 1

n != 7A (where != is not equal to)

Consider n=6 and answer option (A) m=2

6^2 = 36 = 7*5 + 1

Why not (A) then?

Rgds,
TGC!
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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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New post 09 Aug 2013, 05:13
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hi everyone,

I solved this question like this , as 7 is a prime no and n can only take values which are not multiples of 7 so n and 7 will be co prime hence as per fermat little theorem value of m will be 7-1 = 6 for any value of n which are not multiples of n

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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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New post 08 Sep 2013, 08:55
abhishekkhosla wrote:
hi everyone,

I solved this question like this , as 7 is a prime no and n can only take values which are not multiples of 7 so n and 7 will be co prime hence as per fermat little theorem value of m will be 7-1 = 6 for any value of n which are not multiples of n


for those who know fermat little theorem(s) its good but for others, plugging in will always help as explained by Sudhir... thanks Abhishek though for reminding this theorem

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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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AnkitK 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


Responding to a pm:

It is a conceptual question and it easy to figure out if you understand binomial theorem discussed here: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

n^m leaves a remainder of 1 after division by 7 for all n (obviously n cannot be a multiple of 7 because that would leave a remainder of 0)
So n can be of the form (7a + 1) or (7a + 2) or (7a + 3) or (7a + 4) or (7a +5) or (7a + 6)

According to binomial, the remainder I will get when I divide n^m by 7 will depend on the last term i.e. 1^m or 2^m or 3^m or 4^m or 5^m or 6^m.
We need a value of m such that when we divide any one of these 6 terms by 7, we always get a remainder 1.

Can m be 2? 1^2 leaves remainder 1 but 2^2 leaves remainder 4. So no
Can m be 3? 1^3 leaves remainder 1, 2^3 leaves remainder 1. 3^3 leaves remainder 6. So no
Can m be 4? 1^4 leaves remainder 1 but 2^4 leaves remainder 2. So no
Can m be 5? 1^5 leaves remainder 1 but 2^5 leaves remainder 4. So no'

m must be 6 because that is the only option left.
1^6 leaves remainder 1, 2^6 leaves remainder 1 etc. So we see that we are correct.
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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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New post 18 Sep 2013, 02:14
VeritasPrepKarishma wrote:
AnkitK 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


Responding to a pm:

It is a conceptual question and it easy to figure out if you understand binomial theorem discussed here: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

n^m leaves a remainder of 1 after division by 7 for all n (obviously n cannot be a multiple of 7 because that would leave a remainder of 0)
So n can be of the form (7a + 1) or (7a + 2) or (7a + 3) or (7a + 4) or (7a +5) or (7a + 6)

According to binomial, the remainder I will get when I divide n^m by 7 will depend on the last term i.e. 1^m or 2^m or 3^m or 4^m or 5^m or 6^m.
We need a value of m such that when we divide any one of these 6 terms by 7, we always get a remainder 1.

Can m be 2? 1^2 leaves remainder 1 but 2^2 leaves remainder 4. So no
Can m be 3? 1^3 leaves remainder 1, 2^3 leaves remainder 1. 3^3 leaves remainder 6. So no
Can m be 4? 1^4 leaves remainder 1 but 2^4 leaves remainder 2. So no
Can m be 5? 1^5 leaves remainder 1 but 2^5 leaves remainder 4. So no'

m must be 6 because that is the only option left.
1^6 leaves remainder 1, 2^6 leaves remainder 1 etc. So we see that we are correct.


I understand your solution, But how is it associated with Binomial theorem, we have solved it with plugin here?
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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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New post 18 Sep 2013, 07:30
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honchos wrote:
VeritasPrepKarishma wrote:
AnkitK 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


Responding to a pm:

It is a conceptual question and it easy to figure out if you understand binomial theorem discussed here: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

n^m leaves a remainder of 1 after division by 7 for all n (obviously n cannot be a multiple of 7 because that would leave a remainder of 0)
So n can be of the form (7a + 1) or (7a + 2) or (7a + 3) or (7a + 4) or (7a +5) or (7a + 6)

According to binomial, the remainder I will get when I divide n^m by 7 will depend on the last term i.e. 1^m or 2^m or 3^m or 4^m or 5^m or 6^m.
We need a value of m such that when we divide any one of these 6 terms by 7, we always get a remainder 1.

Can m be 2? 1^2 leaves remainder 1 but 2^2 leaves remainder 4. So no
Can m be 3? 1^3 leaves remainder 1, 2^3 leaves remainder 1. 3^3 leaves remainder 6. So no
Can m be 4? 1^4 leaves remainder 1 but 2^4 leaves remainder 2. So no
Can m be 5? 1^5 leaves remainder 1 but 2^5 leaves remainder 4. So no'

m must be 6 because that is the only option left.
1^6 leaves remainder 1, 2^6 leaves remainder 1 etc. So we see that we are correct.


I understand your solution, But how is it associated with Binomial theorem, we have solved it with plugin here?


The remainder you get when you divide (7a + 1)^m by 7 will be 1. The remainder you get when you divide (7a + 2)^m by 7 is determined by 2^m. This is determined by binomial theorem. The link explains you why.
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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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New post 10 Jul 2015, 10:26
TGC wrote:
Somehow didn't get this question as per below.

n^m = 7Q + 1

n != 7A (where != is not equal to)

Consider n=6 and answer option (A) m=2

6^2 = 36 = 7*5 + 1

Why not (A) then?

Rgds,
TGC!


Can anyone please help to explain this issue! I have the same question!
Another case can be: n=2, m=3: 2^3 = 8 = 7*1 + 1 (n^m = 7*a + 1)

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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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New post 14 Jul 2015, 01:41
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Beat720 wrote:
TGC wrote:
Somehow didn't get this question as per below.

n^m = 7Q + 1

n != 7A (where != is not equal to)

Consider n=6 and answer option (A) m=2

6^2 = 36 = 7*5 + 1

Why not (A) then?

Rgds,
TGC!


Can anyone please help to explain this issue! I have the same question!
Another case can be: n=2, m=3: 2^3 = 8 = 7*1 + 1 (n^m = 7*a + 1)


The question says that remainder should be 1 for all values of n. So n could be 1 or 2 or 3 or 4 etc, remainder when n^m is divided by 7 will ALWAYS be 1. Check for a few values of n.

In case m = 2,
1^2 = 1 - when 1 is divided by 7, remainder is 1 - fine
2^2 = 4 - when 4 is divided by 7, remainder is 4 - not acceptable

In case m = 3,
1^3 = 1 - when 1 is divided by 7, remainder is 1 - fine
2^3 = 8 - when 8 is divided by 7, remainder is 1 - fine
3^3 = 27 - when 27 is divided by 7, remainder is 6 - not acceptable

Only in case m = 6, for every value of n, you will get remainder 1.

Answer (E)

Another thing, if m = 2 or m = 3 were the answer, m = 6 would automatically be the answer too because

n^6 = (n^3)^2 = (n^2)^3

But in problem solving questions, you have only one correct answer.
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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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New post 20 Aug 2016, 10:58
AnkitK 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


n^m = 7x +1 ; numbers are 8,15,22,29,36,43,50,57,64,..

The squares we see are : 36 = 3^2x2^2 -> cannot be expressed as n^m
Next
64 =2^6 -> expressed in n^m
So 6 is answer. E.

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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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New post 30 Aug 2017, 00:05
Option E satisfies the required condition (remainder is 1 when n^m is divided by 7) for all values of 'n' whereas option B satisfies it only when n=2 (remainder is 1 when 2^3 is divided by 7 but it is 6 when n=3: 3^3=27/7). Actually, the question itself is confusing. It should have stipulated: "m must be (instead of, could be) equal to:".

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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]

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New post 14 Sep 2017, 14:33
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AnkitK 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


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:
[Reveal] Spoiler:
E


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Re: If n^m leaves a remainder of 1 after division by 7 for all p   [#permalink] 14 Sep 2017, 14:33
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