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Bunuel
When n is divided by 6, the remainder is 2. What is the remainder when (n + 3)^2 is divided by 12?

A. 1
B. 2
C. 3
D. 4
E. 5

Least possible value of n = 8

\((n + 3)^2\) = \((8 + 3)^2\) =>121

121 divided by 12 will have remainder 1, so answer will be (A)

Hi,
you are correct in answer and approach BUT least possible value of n is not 8..
positive least value is 2..
and with no restrictions it can go to anything example : -58 = 6*-10 +2.. here remainder is 2 again..
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n=6q+2 where q is a positive integer.

Also, (n+3)^2 when divided by 12 leaves a remainder, say r
Therefore, (n+3)^2
= (6q+2+3)^2
= (6q+5)^2
= 36q^2+25+60q

Now, consider each of the terms above separately to see what is the remainder when they are divided by 12.

36q^2 is divisible by 12 as 36 is divisible by 12. Hence remainder is 0
60q is divisible by 12 as 60 is divisible by 12.Hence remainder is 0
25 when divided by 12 leaves a remainder 1

Hence, remainder for (n+3)^2 is 1
Hence correct option is A
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N=8,14,20. (N+3)^2= 121,289,529. In all the cases the remainder is 1.
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When n is divided by 6, the remainder is 2. What is the remainder when (n + 3)^2 is divided by 12?

A. 1
B. 2
C. 3
D. 4
E. 5

n = 2, 8, 14 etc.

Therefore we can start with n as "2", (2 + 3) ^2 = 25

Now 25/12 gives us remainder = 1

(A).
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Bunuel
When n is divided by 6, the remainder is 2. What is the remainder when (n + 3)^2 is divided by 12?

A. 1
B. 2
C. 3
D. 4
E. 5

Given: When n is divided by 6, the remainder is 2.

Asked: What is the remainder when (n + 3)^2 is divided by 12?

nmod6 = 2
n = 6k + 2
n+3 = 6k + 2+3 = 6k + 5

(n+3)^2mod12 = (6k+5)^2mod12 = (36k^2 + 60k + 25)mod12 = 25mod12 = 1mod12

IMO A
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