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If N > 4 and N is a prime number, then what is the remainder when N

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If N > 4 and N is a prime number, then what is the remainder when N  [#permalink]

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New post 17 Aug 2018, 02:43
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If N > 4 and N is a prime number, then what is the remainder when N  [#permalink]

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New post 17 Aug 2018, 03:01
Bunuel wrote:
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


OA: B

As \(N\) can be any prime number such that \(N>4\) , Taking \(N =5\) and squaring it, we get \(N^2=25.\)
\(\frac{25}{6}\) will leave \(1\) as remainder.
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If N > 4 and N is a prime number, then what is the remainder when N  [#permalink]

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New post 17 Aug 2018, 04:53
All Prime numbers greater than 3 can be expressed in the form of 6k+1 or 6k-1 , where k is a not negative integer.

Let 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.

B is the answer

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If N > 4 and N is a prime number, then what is the remainder when N  [#permalink]

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New post 17 Aug 2018, 05:30
Bunuel wrote:
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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Re: If N > 4 and N is a prime number, then what is the remainder when N  [#permalink]

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New post 17 Aug 2018, 07:28
Bunuel wrote:
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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Re: If N > 4 and N is a prime number, then what is the remainder when N  [#permalink]

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New post 17 Aug 2018, 08:15
Bunuel wrote:
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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Re: If N > 4 and N is a prime number, then what is the remainder when N  [#permalink]

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New post 18 Aug 2018, 09:39
if N = 5
then N^2 = 25
25/6 remainder is 1

So Answer is B
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Re: If N > 4 and N is a prime number, then what is the remainder when N  [#permalink]

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New post 18 Aug 2018, 16:46
Bunuel wrote:
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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Re: If N > 4 and N is a prime number, then what is the remainder when N  [#permalink]

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New post 20 Aug 2018, 11:20
Bunuel wrote:
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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Re: If N > 4 and N is a prime number, then what is the remainder when N   [#permalink] 20 Aug 2018, 11:20
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