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If n is a prime number greater than 3, what is the remainder

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If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post Updated on: 24 Apr 2012, 12:46
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If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12?

A. 0
B. 1
C. 2
D. 3
E. 5

it's a simple quesiton, but the solutuion is inspiring.

Spoiler: :: Solution
n^2-1=(n-1)(n+1)
since (n-1) and (n+1) are consecutive even numbers,one of them can be divided by 2, another one can be divided by 4;
and because n can not be divided by 3, so one of (n-1) and (n+1) can be divided by 3.
So (n-1)(n+1)=n^2-1 is divisible by 24, then the remainder of n^2 divided by 24 is 1.

Originally posted by chonepiece on 28 Oct 2011, 05:13.
Last edited by Bunuel on 24 Apr 2012, 12:46, edited 2 times in total.
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 01 Nov 2012, 05:54
8
5
sachindia wrote:
VeritasPrepKarishma wrote:
sdas wrote:
all i did was pick numbers and try...got remainder as 1 always. so B. anything wrong in my approach pls advice


No. Nothing wrong. Just that you don't need to try many numbers. There will only be one answer to a PS question. So all you need to do is try any one number greater than 3. Whatever you get, that will be the answer in every case.


I wonder why every prime no greater than 3 when squared and divided by 4 results in the remainder of 1 :shock:


Any prime number \(p\) greater than 3 could be expressed as \(p=6n+1\) or \(p=6n+5\) (\(p=6n-1\)), where \(n\) is an integer >1.

That's because any prime number \(p\) greater than 3 when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).

But:
Note that, not all number which yield a remainder of 1 or 5 upon division by 6 are primes, so vise-versa of above property is not correct. For example 25 (for \(n=4\)) yields a remainder of 1 upon division by 6 and it's not a prime number.

Now, if a prime is of the form \(p=6n+1\), then \(p^2=36n^2+12n+1=12(3n^2+n)+1\) and if a prime is of the form \(p=6n-1\), then \(p^2=36n^2-12n+1=12(3n^2-n)+1\). Both yield the remainder of 1 when divided by 12.

Hope it's clear.
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Re: A question with inspiring solution  [#permalink]

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New post 28 Oct 2011, 06:25
3
chonepiece wrote:
If n is a prime number greater than 3, what is the remainder when n is divided by 12?
0
1
2
3
5


This question is wrong. Please correct it.

what is the remainder when n is divided by 12
should be
what is the remainder when n^2 is divided by 12
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 09 Feb 2012, 14:57
3
1
chonepiece wrote:
If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12?
A. 0
B. 1
C. 2
D. 3
E. 5

it's a simple quesiton, but the solutuion is inspiring.

Spoiler: :: Solution
n^2-1=(n-1)(n+1)
since (n-1) and (n+1) are consecutive even numbers,one of them can be divided by 2, another one can be divided by 4;
and because n can not be divided by 3, so one of (n-1) and (n+1) can be divided by 3.
So (n-1)(n+1)=n^2-1 is divisible by 24, then the remainder of n^2 divided by 24 is 1.


There are several algebraic ways to solve this question including the one under the spoiler. But the easiest way is as follows: since we cannot have two correct answers just pick a prime greater than 3, square it and see what would be the remainder upon division of it by 12.

n=5 --> n^2=25 --> remainder upon division 25 by 12 is 1.

Answer: B.
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 10 Feb 2012, 02:33
3
3
chonepiece wrote:
If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12?
A. 0
B. 1
C. 2
D. 3
E. 5

it's a simple quesiton, but the solutuion is inspiring.

Spoiler: :: Solution
n^2-1=(n-1)(n+1)
since (n-1) and (n+1) are consecutive even numbers,one of them can be divided by 2, another one can be divided by 4;
and because n can not be divided by 3, so one of (n-1) and (n+1) can be divided by 3.
So (n-1)(n+1)=n^2-1 is divisible by 24, then the remainder of n^2 divided by 24 is 1.


Check out these posts for more on divisibility of consecutive integers:
http://www.veritasprep.com/blog/2011/09 ... c-or-math/
http://www.veritasprep.com/blog/2011/09 ... h-part-ii/

and of course, the most efficient solution would be what Bunuel suggested - Pick a prime number > 3 and check for it!
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 10 Feb 2012, 02:47
1
all i did was pick numbers and try...got remainder as 1 always. so B. anything wrong in my approach pls advice
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 10 Feb 2012, 02:54
1
sdas wrote:
all i did was pick numbers and try...got remainder as 1 always. so B. anything wrong in my approach pls advice


No. Nothing wrong. Just that you don't need to try many numbers. There will only be one answer to a PS question. So all you need to do is try any one number greater than 3. Whatever you get, that will be the answer in every case.
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 10 Feb 2012, 03:41
cool karishma. i only tried 5^2 and 7^2 both i got 1 as remainders
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Number properties  [#permalink]

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New post 23 Apr 2012, 06:45
Hi

This is my first post, and am hoping someone can help me out with this question.

If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12

A) 0
B) 1
C) 2
D) 3
E) 5

What I am looking for is advice on how to approach this problem, what are the math rules I can apply.

Many thanks
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 24 Apr 2012, 10:36
I tried solving it may be a lengthier method .. prime number greater than 6 is 6k+1 or 6k-1.. thus squaring say 6k+1 will give.. 36k2 + 12K + 1.... remainder 1... to double check for 5... 25/12..remainder 1... thus B.
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 07 May 2012, 17:02
Bunuel wrote:
chonepiece wrote:
If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12?
A. 0
B. 1
C. 2
D. 3
E. 5

it's a simple quesiton, but the solutuion is inspiring.

Spoiler: :: Solution
n^2-1=(n-1)(n+1)
since (n-1) and (n+1) are consecutive even numbers,one of them can be divided by 2, another one can be divided by 4;
and because n can not be divided by 3, so one of (n-1) and (n+1) can be divided by 3.
So (n-1)(n+1)=n^2-1 is divisible by 24, then the remainder of n^2 divided by 24 is 1.


There are several algebraic ways to solve this question including the one under the spoiler. But the easiest way is as follows: since we can not have two correct answers just pick a prime greater than 3, square it and see what would be the remainder upon division of it by 12.

n=5 --> n^2=25 --> remainder upon division 25 by 12 is 1.

Answer: B.

Hello
if n is a prime number bigger than 3 it can also be 7 right ?
Hence 7 ^2 =49/12 Is not equal to 1
can I pick 7 or ?

thanks

best regards
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 08 May 2012, 01:18
1
keiraria wrote:
Bunuel wrote:
chonepiece wrote:
If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12?
A. 0
B. 1
C. 2
D. 3
E. 5

it's a simple quesiton, but the solutuion is inspiring.

Spoiler: :: Solution
n^2-1=(n-1)(n+1)
since (n-1) and (n+1) are consecutive even numbers,one of them can be divided by 2, another one can be divided by 4;
and because n can not be divided by 3, so one of (n-1) and (n+1) can be divided by 3.
So (n-1)(n+1)=n^2-1 is divisible by 24, then the remainder of n^2 divided by 24 is 1.


There are several algebraic ways to solve this question including the one under the spoiler. But the easiest way is as follows: since we can not have two correct answers just pick a prime greater than 3, square it and see what would be the remainder upon division of it by 12.

n=5 --> n^2=25 --> remainder upon division 25 by 12 is 1.

Answer: B.

Hello
if n is a prime number bigger than 3 it can also be 7 right ?
Hence 7 ^2 =49/12 Is not equal to 1
can I pick 7 or ?

thanks

best regards


As explained. you can pick ANY prime greater than 3.

Also, 49 divided by 12 yields the remainder of 1: 49=4*12+1.

Hope it's clear.
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 01 Nov 2012, 03:13
VeritasPrepKarishma wrote:
sdas wrote:
all i did was pick numbers and try...got remainder as 1 always. so B. anything wrong in my approach pls advice


No. Nothing wrong. Just that you don't need to try many numbers. There will only be one answer to a PS question. So all you need to do is try any one number greater than 3. Whatever you get, that will be the answer in every case.


I wonder why every prime no greater than 3 when squared and divided by 4 results in the remainder of 1 :shock:
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 02 Nov 2012, 01:11
Do we really need to know this concept for GMAT :shock:
I find this going slighty above my head. . :roll:
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 17 Nov 2013, 13:39
If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12?

A. 0
B. 1
C. 2
D. 3
E. 5

whenever any square of prime number>2 is divisible by 12 the remainder is always 1.
Check through options:

here in case n = 5,7,11 and so on.

5^2 = 25 when its divisible by 12,the remainder is 1.
7^2 = 49 when its divisible by 12, still the remainder is 1.
11^2 = 121 when its divisible by 12,the remaider is 1.

So answer of this question is (b).
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 01 May 2014, 03:13
pavanpuneet wrote:
I tried solving it may be a lengthier method .. prime number greater than 6 is 6k+1 or 6k-1.. thus squaring say 6k+1 will give.. 36k2 + 12K + 1.... remainder 1... to double check for 5... 25/12..remainder 1... thus B.



Too much of an overkill. As Bunuel suggested , here the best approach would be to put values and get the response.
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 13 Feb 2015, 21:17
Hello,

In OG explanation it is mentioned that

Consider n2/12 as each n divided by 6 ????

Well I don't think that's correct. It should be read as n2 divided by 6 X 2... Please classify...

Thanks

Posted from my mobile device
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 16 Feb 2015, 03:41
vikasbansal227 wrote:
Hello,

In OG explanation it is mentioned that

Consider n2/12 as each n divided by 6 ????

Well I don't think that's correct. It should be read as n2 divided by 6 X 2... Please classify...

Thanks

Posted from my mobile device


Below is a solution from OG:

The simplest way to solve this problem is to choose a prime number greater than 3 and divide its square by 12 to see what the remainder is. For example, if n= 5, then n^2 =25, and the remainder is 1 when 25 is divided by 12. A second prime number can be used to check the result. For example, if n = 7, then n^2 =49, and the remainder is 1 when 49 is divided by 12. Because only one of the answer choices can be correct, the remainder must be 1.

For the more mathematically inclined, consider the remainder when each prime number n greater than 3 is divided by 6. The remainder cannot be 0 because that would imply that n is divisible by 6, which is impossible since n is a prime number. The remainder cannot be 2 or 4 because that would imply that n is even, which is impossible since n is a prime number greater than 3. The remainder cannot be 3 because that would imply that n is divisible by 3, which is impossible since n is a prime number greater than 3. Therefore, the only possible remainders when a prime number n greater than 3 is divided by 6 are 1 and 5. Thus, n has the form 6q + 1 or 6q + 5, where q is an integer, and, therefore, n^2 has the form 36q^2 +12q +1 =12(3q^2 +q) +1 or 36q^2+60q +25 =12(3q^2+5q +2)+ 1. In either case, n^2 has a remainder of 1 when divided by 12.


Can you please tell me which part there is not clear? Thank you.
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 04 May 2016, 09:52
chonepiece wrote:
If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12?

A. 0
B. 1
C. 2
D. 3
E. 5



Solution:

We see that n can be ANY PRIME NUMBER GREATER THAN 3. Let’s choose the smallest prime number greater than 3 and substitute it for n; that number is 5.

We know that 5 squared is 25, so we now divide 25 by 12:

25/12 = 2, Remainder 1.

If you are not convinced by trying just one prime number, try another one. Let’s try 7. We know that 7 squared equals 49, so we now divide 49 by 12:

49/12 = 4, Remainder 1.

It turns out that in this problem it doesn’t matter which prime number (greater than 3) we choose. The remainder will always be 1 when its square is divided by 12.

Answer B.
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Re: If n is a prime number greater than 3, what is the remainder   [#permalink] 04 May 2016, 09:52

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