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Bunuel
If N > 4 and N is a prime number, then what is the remainder when N^2 is divided by 6?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Smart Question:

It can be looked ta in two ways

1) By Properties of Prime numbers Every prime number greater than 3 when divided by 6 leaves remainder either +1 or -1
i.e. N^2 divided by 6 will eave remainder (+1)^2 i.e. +1 hence

Answer: option B

2) Take a value of a few prime numbers greater than 4
i.e. N may be 5, 7, 11, 13, ... etc
i.e. N^2 may be 25, 49, 121, 169, ... etc

Dividing each of these values of N^2 by 6 leaves remainder 1 hence

Answer: option B
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Bunuel
If N > 4 and N is a prime number, then what is the remainder when N^2 is divided by 6?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Possible Values of \(N = { 5, 7 , 11 , 13..............}\)

Try with the possible values of \(\frac{N^2}{6}\) , in each case answer must be 1, Thus answer must be (B) 1
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Bunuel
If N > 4 and N is a prime number, then what is the remainder when N^2 is divided by 6?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

I think the easiest way to solve this qs is number plugging. Since we can't have 2 correct answers at the same time, we just need to pick a number and see which answer choice matches the result.
In this case let N=5 --> \(N^2\) = 25 --> remainder when 25 is divided by 6 is 1.
Answer B.
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if N = 5
then N^2 = 25
25/6 remainder is 1

So Answer is B
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Bunuel
If N > 4 and N is a prime number, then what is the remainder when N^2 is divided by 6?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


we can plug numbers for N. It's given that N is prime greater than 4.

So, N can be : 5, 7 , 11.

\(\frac{(5)^2}{6}= 24 + 1\)

\(\frac{(7)^2}{6}= 48 + 1\)

So, in all case we will have a remainder 1.

The best answer is B.
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Bunuel
If N > 4 and N is a prime number, then what is the remainder when N^2 is divided by 6?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

We can let N = 5, so N^2 = 25, and we have:

25/6 = 4 remainder 1

Answer: B
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