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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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03 Dec 2016, 23:42
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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04 Dec 2016, 00:08
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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30 Dec 2016, 09:03
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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15 Mar 2018, 05:17
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, 05:17

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