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

[BTGModeratorVI]

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

Answer: B
Source: Official guide

[Jay@ManhattanReview]

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

Answer: B
Source: Official guide

Say n = 5

So n^2 = 5^2 = 25

So, the remainder when n^2 = 25 is divided by 12 = 1

The correct answer: B

Hope this helps!

-Jay
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[Brent@GMATPrepNow]

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

Answer: B
Source: Official guide

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 = 5² = 25, and when we divide 25 by 12, we get 2 with REMAINDER 1

Answer: B

Cheers,
Brent

[Scott@TargetTestPrep]

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

Answer: B
Source: Official guide

Solution:

We can let n = 5. Thus n^2 = 25 and 25/12 = 2 R 1.

(Note: The answer choices don’t have a choice such as “Can’t be determined.” The question is not asking “what could be the remainder” either. We can safely say the correct answer must be 1, though we only used one value for n. If we want to further ensure that the answer must be 1, we can use another value for n, such as n = 7. We see that n^2 = 49 and 49/12 = 4 R 1. The remainder once again is 1.)

Alternate Solution:

Notice that n^2 - 1 is divisible by 4 since n^2 - 1 = (n - 1)(n + 1) and both n - 1 and n + 1 are even numbers (as n^2 - 1 has at least two factors of 2, it will be divisible by 4).

Further, n - 1, n and n + 1 are three consecutive integers; thus exactly one of them is divisible by 3. Since n is prime, n cannot be divisible by 3 (as n > 3); so either n - 1 or n + 1 must be divisible by 3.

Since n^2 - 1 is divisible by both 3 and 4, it is divisible by LCM(3, 4) = 12. Since n^2 - 1 is divisible by 12, n^2 leaves a remainder of 1 when divided by 12.

Answer: B