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

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Joined: 10 Aug 2011
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If n is a prime number greater than 3, what is the remainder when n^2  [#permalink]

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Updated on: 13 Aug 2019, 02:38
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Question Stats:

95% (00:43) correct 5% (01:03) wrong based on 1607 sessions

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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

Originally posted by chonepiece on 28 Oct 2011, 05:13.
Last edited by Bunuel on 13 Aug 2019, 02:38, edited 3 times in total.
Edited the question
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If n is a prime number greater than 3, what is the remainder when n^2  [#permalink]

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26 Aug 2012, 02:56
8
18
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

There are several algebraic ways to solve this question, 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.

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

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27 Jun 2013, 10:00
34
13
Bunuel 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

Practice Questions
Question: 26
Page: 155
Difficulty: 600

All Prime numbers greater than 3 and upto 1000000000000000 can be expressed in the form of 6k+1 or 6k-1 , where k is a not negative integer.

Say N = 6k+1
N^2 = (6k+1)^2 = 36K^2 + 12K + 1 = 12(3K^2 + K) +1

Since 12(3K^2+K) is exactly divisible by 12 , therefore N^2 when divided by 12 leaves a remainder as 1.

Same can be proved for N = 6K -1
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Re: If n is a prime number greater than 3, what is the remainder when n^2  [#permalink]

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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 when n^2  [#permalink]

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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 when n^2  [#permalink]

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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 when n^2  [#permalink]

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

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01 Nov 2012, 05:54
8
6
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

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: If n is a prime number greater than 3, what is the remainder when n^2  [#permalink]

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02 Jul 2013, 10:16
All the prime numbers >2 end with 1,3,5,7. Since the question stem says Primes >3, we need not check those ending with 1 & 3. so pick any numbers ending with 5&7 , you will get the same remainder.
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Re: If n is a prime number greater than 3, what is the remainder when n^2  [#permalink]

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02 Jul 2013, 10:41
2
pavan2185 wrote:
All the prime numbers >2 end with 1,3,5,7.

No prime no can ever end in 5, except 5.

Since the question stem says Primes >3, we need not check those ending with 1 & 3.

11,13 are primes, which are greater than 3 and end with 1 and 3 respectively.

Moreover,the question asks for the remainder when $$n^2$$ is divided by 12, not n.
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Re: If n is a prime number greater than 3, what is the remainder when n^2  [#permalink]

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02 Jul 2013, 10:51
2
mau5 wrote:
pavan2185 wrote:
All the prime numbers >2 end with 1,3,5,7.

No prime no can ever end in 5, except 5.

Since the question stem says Primes >3, we need not check those ending with 1 & 3.

11,13 are primes, which are greater than 3 and end with 1 and 3 respectively.

Moreover,the question asks for the remainder when $$n^2$$ is divided by 12, not n.

We can simply check for the first prime which is greater than 3, so for 5: n=5 --> n^2=25 --> remainder upon division 25 by 12 is 1.

Also, primes greater than 5 can have the units digit of 1, 3, 7, OR 9.
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Re: If n is a prime number greater than 3, what is the remainder when n^2  [#permalink]

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25 Dec 2013, 11:39
koolgmat wrote:
Bunuel 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

Practice Questions
Question: 26
Page: 155
Difficulty: 600

All Prime numbers greater than 3 and upto 1000000000000000 can be expressed in the form of 6k+1 or 6k-1 , where k is a not negative integer.

Say N = 6k+1
N^2 = (6k+1)^2 = 36K^2 + 12K + 1 = 12(3K^2 + K) +1

Since 12(3K^2+K) is exactly divisible by 12 , therefore N^2 when divided by 12 leaves a remainder as 1.

Same can be proved for N = 6K -1

How do we know that the prime number in questions is not larger than 10^15?

So that we know how to, how do we solve this problem algebraically?
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Re: If n is a prime number greater than 3, what is the remainder when n^2  [#permalink]

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26 Dec 2013, 03:39
bschoolaspirant9 wrote:
koolgmat wrote:
Bunuel 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

Practice Questions
Question: 26
Page: 155
Difficulty: 600

All Prime numbers greater than 3 and upto 1000000000000000 can be expressed in the form of 6k+1 or 6k-1 , where k is a not negative integer.

Say N = 6k+1
N^2 = (6k+1)^2 = 36K^2 + 12K + 1 = 12(3K^2 + K) +1

Since 12(3K^2+K) is exactly divisible by 12 , therefore N^2 when divided by 12 leaves a remainder as 1.

Same can be proved for N = 6K -1

How do we know that the prime number in questions is not larger than 10^15?

So that we know how to, how do we solve this problem algebraically?

n^2 can be greater than 10^15 and the property would still hold true. koolgmat incorrectly limited the upper limit to 10^15.

The property koolgmat is referring to is: ANY prime number $$p$$ greater than 3 can 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.

Hope it's clear.

P.S. Similar question to practice: if-n-4p-where-p-is-a-prime-number-greater-than-2-how-man-144781.html
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Re: If n is a prime number greater than 3, what is the remainder when n^2  [#permalink]

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31 Mar 2015, 12:07
Bunuel 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

Practice Questions
Question: 26
Page: 155
Difficulty: 600

There are infinite prime numbers greater than 3.
So one this is sure that all prime numbers (greater than 3) when squared and then divided by 12 will have the same remainder.
So let us assume that the prime number is 5.
So, 5^2 = 25 when divided by 12 gives 1 as remainder.
Hence option (B).

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

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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.

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

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Re: If n is a prime number greater than 3, what is the remainder when n^2   [#permalink] 30 Apr 2019, 01:52
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