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What is the remainder when n2 + 7 is divided by 8, where n is an odd p

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What is the remainder when n2 + 7 is divided by 8, where n is an odd p  [#permalink]

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New post 19 Dec 2018, 11:21
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Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p  [#permalink]

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New post 19 Dec 2018, 12:26
EgmatQuantExpert wrote:
What is the remainder when \(n^2 + 7\) is divided by 8, where n is an odd prime number?

A. 0
B. 2
C. 3
D. 4
E. 6


let n=3
3^2+7=16
16/8 gives 0 remainder
A
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Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p  [#permalink]

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New post 20 Dec 2018, 01:30
upon checking relation for n^2+7 ; at n= 1,3,5,7,11.. we can say that the remainder would always be 0

IMO A ..


EgmatQuantExpert wrote:
What is the remainder when \(n^2 + 7\) is divided by 8, where n is an odd prime number?

    A. 0
    B. 2
    C. 3
    D. 4
    E. 6

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Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p  [#permalink]

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New post 01 Jan 2019, 22:46
1

Solution


Given:
We are given that
    • “n” is an odd prime number

To find:
We need to find,
    • The remainder when \(n^2 + 7\) is divided by 8

Approach and Working:
    • If we observe, all the positive integers can be written in the form 4k or + 4k + 1 or 4k + 2 or 4k + 3, where k is a non-negative integer.
      o Out of these, 4k and 4k + 2 cannot be prime numbers, since they are divisible by 2, for any value of k.

    • Thus, all odd prime numbers must be of the form 4k + 1 or 4k + 3

So, \(n^2\) will be \((4k + 1)^2\) or \((4k + 3)^2\)
    • And, \(n^2 + 7\) = \((16k^2 + 8k + 8)\) or \((16k^2 + 24k + 16)\) = \(8(2k^2 + k + 1)\) or \(8(2k^2 + 3k + 2)\)
    • Thus, both these expressions are divisible by 8, for any value of k

Therefore, the remainder is 0

Hence the correct answer is Option A.

Answer: A

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What is the remainder when n2 + 7 is divided by 8, where n is an odd p  [#permalink]

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New post 28 Jan 2019, 01:03
The question is simple with substitution. The solution above seems to choose 4 arbitrarily.
Why not all positive integers can be expressed 2k, 2k+1 resulting in 4(k^2+k+2)?
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Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p  [#permalink]

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New post 04 Mar 2019, 18:57
philipssonicare wrote:
The question is simple with substitution. The solution above seems to choose 4 arbitrarily.
Why not all positive integers can be expressed 2k, 2k+1 resulting in 4(k^2+k+2)?


I concur.. Infact when we take 4k+1 we miss out on 3 which is a possible scenario which is not the case while taking 2k+1..
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Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p  [#permalink]

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New post 04 Mar 2019, 19:01
EgmatQuantExpert wrote:

Solution


Given:
We are given that
    • “n” is an odd prime number

To find:
We need to find,
    • The remainder when \(n^2 + 7\) is divided by 8

Approach and Working:
    • If we observe, all the positive integers can be written in the form 4k or + 4k + 1 or 4k + 2 or 4k + 3, where k is a non-negative integer.
      o Out of these, 4k and 4k + 2 cannot be prime numbers, since they are divisible by 2, for any value of k.

    • Thus, all odd prime numbers must be of the form 4k + 1 or 4k + 3

So, \(n^2\) will be \((4k + 1)^2\) or \((4k + 3)^2\)
    • And, \(n^2 + 7\) = \((16k^2 + 8k + 8)\) or \((16k^2 + 24k + 16)\) = \(8(2k^2 + k + 1)\) or \(8(2k^2 + 3k + 2)\)
    • Thus, both these expressions are divisible by 8, for any value of k

Therefore, the remainder is 0

Hence the correct answer is Option A.

Answer: A

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When we take 4k+1 aren't we missing out on 3 which is a possible scenario??
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What is the remainder when n2 + 7 is divided by 8, where n is an odd p  [#permalink]

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New post 04 Mar 2019, 21:26
Quote:
The question is simple with substitution. The solution above seems to choose 4 arbitrarily.
Why not all positive integers can be expressed 2k, 2k+1 resulting in 4(k^2+k+2)?

We are free to take 2k and 2k + 1 form also. As you mentioned, \((2k + 1)^2 + 7\) will be equal to \(4k^2 + 4k + 8 = 4(k^2 + k) + 8\)
Now, for any value of k, \(k^2 + k\) will be either 0 or an even number, hence \(4(k^2 + k)\) will be always divisible by 8.

Quote:

• Thus, all odd prime numbers must be of the form 4k + 1 or 4k + 3

So, \(n^2\) will be \((4k + 1)^2\) or \((4k + 3)^2\)
    • And, \(n^2 + 7\) = \((16k^2 + 8k + 8)\) or \((16k^2 + 24k + 16)\) = \(8(2k^2 + k + 1)\) or \(8(2k^2 + 3k + 2)\)
    • Thus, both these expressions are divisible by 8, for any value of k

Therefore, the remainder is 0

Quote:
When we take 4k+1 aren't we missing out on 3 which is a possible scenario??


If you look at the solution carefully, you will be able to see that we've considered both cases here: the possibility of 4k + 1 as well as the possibility of 4k + 3. The possible scenario of 3 is covered in the case of 4k + 3.

Hope this helps. :)
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Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p  [#permalink]

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New post 07 Mar 2019, 11:18
n is not equal to 2 because n is an odd prime number, n^2 +7 always gives an even number.

Now using n=3,5,7... always results (n^2 +7) = multiples of 8. Hence the remainder of (n^2 +7)/8 is always 0.
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Re: What is the remainder when n2 + 7 is divided by 8, where n is an odd p   [#permalink] 07 Mar 2019, 11:18
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