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

Question

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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gracie · Answer

let n=3
3^2+7=16
16/8 gives 0 remainder
A

Archit3110 · Answer

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 · Answer

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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philipssonicare · Answer

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)?

pankajpatwari · Answer

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..

pankajpatwari · Answer

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

EgmatQuantExpert · Answer

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. 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.

Sateni7628 · Answer

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.

ManifestDreamMBA · Answer

For n=3, remainder is 0
All prime numbers above 3, can be expressed as 6k+1.
(6k+1)^2 + 7 = 36k^2 +12k +1 +7
12k(3k+1) + 8
We need to know the remainder of division of 12k(3k+1) by 8 now
if k is even, say 2p, we have 24p(6p+1). since 24 is divisible by 8, we have 0 remainder
if k is odd, say 2p+1, we have 12k(6p+3+1) = 24k(3p+2). since 24 is divisible by 8, we have 0 remainder (PS: I have intentionally not substituted one of the k, to keep it simple)

So, answer in all scenarios is 0

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

Prep Toolkit

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