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Math Expert V
Joined: 02 Sep 2009
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If n is a prime number greater than 3, what is the remainder  [#permalink]

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17 00:00

Difficulty:   5% (low)

Question Stats: 95% (00:41) correct 5% (01:01) wrong based on 1417 sessions

### HideShow timer Statistics 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

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Math Expert V
Joined: 02 Sep 2009
Posts: 56244
Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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7
2
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.

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

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30
11
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
##### General Discussion
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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2
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]

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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

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

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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.

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]

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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]

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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]

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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]

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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]

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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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Posts: 56244
Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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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]

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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

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

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Bunuel

Hi,

I came across another way to solve the problem but I need you to validate it. It might lead to the correct answer choice just by fluke!

Here it is:

Any positive integers not multiple of 3, when divided by 3, have a remainder of 1 or 2
Any positive odd integers when divided by 2, have a remainder of 1
So, if n is divided by 2 and 3 the remainder will be 1. ( which the only common option among the remainders of 3 and 2).

Does it make any sense?

Thank you!

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.

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

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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

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  [#permalink]

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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

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  [#permalink]

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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.

Alternate Solution :

Option B

Any prime number greater than 3, can be written as (6k +/- 1); for k (integer) >0.
: So as per Q Stem, n will be of the form of (6k +/- 1) AND $$n^2 = (6k +/- 1)^2 = 36k^2 + 1^2 +/- 12k$$
: So, when divided by 12, $$n^2 = 36k^2 + 1^2 +/- 12k$$, leaves 1 as remainder.
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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