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

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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 04 Dec 2016, 00:42
Here is my solution =>
Method 1 ->
Picking a number.
Let n=5
n^2=25
Remainder with 12 => 1
Method 2->
Every prime number greater than 3 can be written as =>
6k+1 or 6k-1

OR
4k+1 or 4k+1

Here Using the first equation => whether its 6k+1 or 6k-1 => the remainder with 12 will always be one.
Hence B

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If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 04 Dec 2016, 01:08
chonepiece wrote:
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


Let \(n = 5\) , so \(n^2 = 25\) & \(\frac{n^2}{12}\) = Remainder 1
Let \(n = 7\) , so \(n^2 = 49\) & \(\frac{n^2}{12}\) = Remainder 1
Let \(n = 11\) , so \(n^2 = 121\) & \(\frac{n^2}{12}\) = Remainder 1
Let \(n = 13\) , so \(n^2 = 169\) & \(\frac{n^2}{12}\) = Remainder 1


Hence answer will be (B) 1


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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 30 Dec 2016, 10:03
n = 5,7,11..
n^2 = 25,49,121
Always reminds 1
B
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Re: If n is a prime number greater than 3, what is the remainder  [#permalink]

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New post 15 Mar 2018, 06:17
chonepiece wrote:
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

it's a simple quesiton, but the solutuion is inspiring.

Spoiler: :: Solution
n^2-1=(n-1)(n+1)
since (n-1) and (n+1) are consecutive even numbers,one of them can be divided by 2, another one can be divided by 4;
and because n can not be divided by 3, so one of (n-1) and (n+1) can be divided by 3.
So (n-1)(n+1)=n^2-1 is divisible by 24, then the remainder of n^2 divided by 24 is 1.


Pick any prime number > 3, and solve.

Take n = 5, then 5^2 = 25/12 gives you remainder "1"

(B)
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Re: If n is a prime number greater than 3, what is the remainder &nbs [#permalink] 15 Mar 2018, 06:17

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