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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]  19 Jun 2011, 09:42
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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]  19 Jun 2011, 10:16
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]  19 Jun 2011, 10:24
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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]  19 Jun 2011, 10:26
Nice explanation Sudhir.
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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]  09 Aug 2013, 04:07
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]  09 Aug 2013, 04:13
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]  08 Sep 2013, 07:55
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]  17 Sep 2013, 21:03
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Expert's post
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Joined: 17 Apr 2013 Posts: 609 Concentration: Entrepreneurship, Leadership Schools: HBS '16 GMAT 1: 710 Q50 V36 GMAT 2: 750 Q51 V41 GPA: 3.3 Followers: 10 Kudos [?]: 111 [0], given: 277 Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink] 18 Sep 2013, 01:14 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? _________________ Like my post Send me a Kudos It is a Good manner. My Debrief: how-to-score-750-and-750-i-moved-from-710-to-189016.html Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1413 Kudos [?]: 7338 [0], given: 186 Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink] 18 Sep 2013, 06:30 Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]  05 Oct 2014, 10:06
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Re: If n^m leaves a remainder of 1 after division by 7 for all p   [#permalink] 05 Oct 2014, 10:06
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