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

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

koolgmat · Accepted Answer

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

Bunuel · Answer

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.

Bunuel · Answer

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.

KarishmaB · Answer

Please check my signature for the link to the relevant blog posts.
...and of course, the most efficient solution would be what Bunuel suggested - Pick a prime number > 3 and check for it!

sdas · Answer

all i did was pick numbers and try...got remainder as 1 always. so B. anything wrong in my approach pls advice

KarishmaB · Answer

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.

Sachin9 · Answer

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

pavan2185 · Answer

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.

mau5 · Answer

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.

Bunuel · Answer

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.

RustyR · Answer

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?

Bunuel · Answer

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

OptimusPrepJanielle · Answer

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). -- Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: https://www.optimus-prep.com/gmat-on-demand-course

ScottTargetTestPrep · Answer

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.

BrentGMATPrepNow · Answer

A nice fast approach is to  TEST  a possible value of n. 
Since n must be a prime number that's greater than 3, let's TEST   n = 5 
 
If   n = 5  , then n² =   5  ² = 25, and when we divide 25 by 12, we get 2 with REMAINDER 1

Answer: B

Cheers,
Brent

MHIKER · Answer

n=5, n^2=25

\frac{25}{12}= 2 \ Remainder \ is 1

The answer is B

100mitra · Answer

Special thankful notes from GMAT Club Quants book
Any prime number greater than 3 could be expressed as (p=6n+1) or p(=6n-1), 
where n is an integer >1,  reminder will be 1

Answer Option : B (1)

BrushMyQuant · Answer

Given that n is a prime number greater than 3 and we need to find what is the remainder when  n^2  is divided by 12

Lets take a prime number greater than 3
=> n = 5

( Watch this video  to know about  Prime Numbers )

n^2  =  5^2  = 25

n^2  i.e. 25 when divided by 12 will give 1 remainder

So,  Answer will be B 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

totamofficiis · Answer

this proof holds true for entire number line. 
lets say n is a prime number.
if we divide n by 6 the for following are the possibilities.

n mod 6 == {0,1,2,3,4,5}

lets say n mod 6 is 2 or 4 then 
n = 6a+2 or 6a+4 hence divisible by 2 thus n can not be a prime. Same proof by negation can be done for 3. 
so n has to be either 6n+1 or 6n-1 format.

n^2 then == 6a^2+2*6a+1 or 6a^2-2*6a+1
all the terms are divisible by 6 except 1.
so answer is 1.

converse is not true btw.

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

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

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