If P is a prime number greater than 5, what is the remainder when P^2

Question

If P is a prime number greater than 5, what is the remainder when P^2 is divided by 8. A) 4 B) 3 C) 2 D) 1 E) Cannot be determined Note => Kudos for an algebraic approach

chetan2u · Accepted Answer

Hi, here is the algebric approach.. these prime numbers are of teh form 6n+1 or 6n-1.. so P= 6n+1.. P^2 = (6n+1)^2 = 36n^2+12n+1 ... Now 36n^2+12n = 4n(9n+3) .... if n is even... 4n will be div by 8.... if n is odd.. 4n will be div by 4 and 9n+3 will become even and be div by 2,hence 4n*(9n+3) will be div by 4*2=8.. so in P^2=36n^2+12n+1 only 1 is left, Remainder = 1..

rohit8865 · Answer

take square of any prime number
remainder will be 1

Ans D

Senthil1981 · Answer

Answer is D: 1

How to solve :
Given, 
 P^2  mod 8 = x,remainder. Then  P^2  - x would be divisible by 8 giving remainder zero.
So  P^2  - x = (P - \sqrt{x} )(P +  \sqrt{x} ) divisible by zero when divided by 8.

Let's consider the choices, 4 would make \sqrt{4} = 2, which would make both factors as odd numbers.
Obviously 3 and 2 are irrational. Left is 1 and when tried P-1 and P+1 would be even and since P is greater than 5, it is the right answer.

stonecold · Answer

Here is one more way to solve this one=>

All prime numbers greater than 2 are odd.
Hence p must be of the form => 2k+1
p^2=> 4k^2+4k+1=> 4k(k+1)+1= 8k'+1
Hence it will always leave a remainder 1 with 8.

Another important takeaway -> 
 For any odd number k => k^2 will always leave a remainder 1 with 8.

kanusha · Answer

P > 5 ( 7,11,13 .. )
p^2 : two odd prime numbers always result odd, not divisible by 8
 \frac{p^2}{8} 
Reminds either 0 or 1
D

Abhishek009 · Answer

Plug in some values and check

Let P = 7, 11 , 13
  
Result will be 1 in each case, thus answer will be (D)

ScottTargetTestPrep · Answer

If we let P = 7, then the remainder when P^2 is divided by 8 is 1 since 49/8 = 6 remainder 1.

If we let P = 11, then the remainder when P^2 is divided by 8 is 1 since 121/8 = 15 remainder 1.

If we let P = 13, then the remainder when P^2 is divided by 8 is 1 since 169/8 = 21 remainder 1.

At this point, you might wonder: is the remainder always 1? If it is, then the answer will be D; otherwise, the answer will be E. Let’s prove it algebraically.

Since any prime number greater than 5 is odd, and an odd number can be written as 4n + 1 or 4n + 3 for some positive integer n, we can express P as P = 4n + 1 or P = 4n + 3.

Case 1: If P = 4n + 1, we have:

P^2 = (4n + 1)^2 = 16n^2 + 8n + 1

We see that the first two terms are divisible by 8; thus, the remainder must be the last term, which is 1.

Case 2: If P = 4n + 3, we have:

P^2 = (4n + 3)^2 = 16n^2 + 24n + 9

We see that the first two terms are divisible by 8 and the last term 9 has a remainder of 1 when it’s divided by 8; thus, the remainder must be 1.

Thus, we see that when the square of a prime (greater than 5) is divided by 8, the remainder will always be 1.

Alternate Solution:

Let’s consider P^2 - 1 = (P - 1)(P + 1).

Since P > 5, both P - 1 and P + 1 are even because P is prime. Moreover, since P - 1 and P + 1 are consecutive even integers, one of them is divisible by 4. Since each of P - 1 and P + 1 are even and since, furthermore, one of them is divisible by 4, their product, which is P^2 - 1, is divisible by 8. Since P^2 - 1 is divisible by 8, P^2 will leave a remainder of 1 when divided by 8.

Answer: D

bumpbot · Answer

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If P is a prime number greater than 5, what is the remainder when P^2

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If P is a prime number greater than 5, what is the remainder when P^2

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