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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] New post 26 Aug 2012, 01:56
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Re: If n is a prime number greater than 3, what is the remainder [#permalink] New post 26 Aug 2012, 01:56
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SOLUTION:

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.

Answer: B.
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Re: If n is a prime number greater than 3, what is the remainder [#permalink] New post 27 Jun 2013, 09:00
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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 [#permalink] New post 02 Jul 2013, 09:51
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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 [#permalink] New post 02 Jul 2013, 09:41
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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 [#permalink] New post 27 Aug 2012, 10:08
Prime nos. n > 3 are 5,7,11...and their squares n^2 are 25, 49, 121...
Remainder of (n^2)/12 is 1.
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Re: If n is a prime number greater than 3, what is the remainder [#permalink] New post 27 Aug 2012, 10:45
Prime numbers greater than 3 are 5,7,.....17 etc

Remainders , when divided by 12;
(5)^2/12--> 1
(7)^2/12-->1
.
.
.
(17)^2/12-->1

B is the right answer
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Re: If n is a prime number greater than 3, what is the remainder [#permalink] New post 31 Aug 2012, 00:47
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SOLUTION:

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.

Answer: B.
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Re: If n is a prime number greater than 3, what is the remainder [#permalink] New post 27 Feb 2013, 23:23
dont use algebric, pick numbers.

I want to follow this posting
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Re: If n is a prime number greater than 3, what is the remainder [#permalink] New post 03 Mar 2013, 05:43
Bah read the question stem carefully. Was doing a cube problem before this and did it with n^3.
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Re: If n is a prime number greater than 3, what is the remainder [#permalink] New post 22 Apr 2013, 22:25
Bunuel wrote:
SOLUTION:

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.

Answer: B.


hi bunuel,

for the above question if N^2 is divided by 3 or 6 or 8 or 24 then also remainder is always 1.

my question is whether these are the only numbers which when divide N^2 gives the constant remainder or there are other numbers also
which when divide a prime^2 (greater than 3) gives a constant remainder.

i know i am out of topic but if possible do reply because knowing these stuff might save some time in exams.

regards
SKM
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Re: If n is a prime number greater than 3, what is the remainder [#permalink] New post 02 Jul 2013, 09:16
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 [#permalink] New post 02 Jul 2013, 10:09
Bunuel wrote:

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.


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.


Sorry for that "5" Thing and I could not exactly write what I was thinking on my mind to solve this question. ( looks like I should sleep now :) )
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Re: If n is a prime number greater than 3, what is the remainder [#permalink] New post 25 Dec 2013, 10:39
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 [#permalink] New post 26 Dec 2013, 02:39
Expert's post
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 [#permalink] New post 10 Sep 2014, 01:30
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



n is a prime number greater than 3. so it has to be 5, 7, 11....

n^2 /12 =

say 25/12 =1
49/12=1
Hence Answer should be 1..
Re: If n is a prime number greater than 3, what is the remainder   [#permalink] 10 Sep 2014, 01:30
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