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06 Nov 2018, 00:18
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If x is a prime number greater than 3, what is the remainder when x^2 is divided by 8 ?
A. 0
B. 1
C. 3
D. 4
E. 5
A. 0
B. 1
C. 3
D. 4
E. 5
Bunuel
If x is a prime number greater than 3, what is the remainder when x^2 is divided by 8 ?
A. 0
B. 1
C. 3
D. 4
E. 5
A. 0
B. 1
C. 3
D. 4
E. 5
x is a prime . x> 3.
x = 5 , 7 ..........
\(5^2 / 8 = 8*3 + 1\)
\(7^2/8 = 6*8 + 1\)
All case remainder is 1.
The best answer is B.
06 Nov 2018, 01:56
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Bunuel
If x is a prime number greater than 3, what is the remainder when x^2 is divided by 8 ?
A. 0
B. 1
C. 3
D. 4
E. 5
A. 0
B. 1
C. 3
D. 4
E. 5
Since this is a PS question, we will have a unique answer here. Whatever is true for all such values for x, will be true for say x = 5 too. So put x = 5
When 5^2 = 25 is divided by 8, remainder = 1
Answer (B)
But this brings an interesting concept forward: For all prime numbers greater than 3, remainder when their square is divided by 8 is 1. Why?
Let's see.
Every prime number greater than 3 is of the form (6a + 1) or (6a - 1).
Case 1: (6a + 1)^2 = 36a^2 + 12a + 1 = 4a(9a + 3) + 1
---- If a is odd, (9a + 3) = Odd + Odd = Even
So 4a(9a + 3) is divisible by 8. Remainder 1
---- If a is even, again 4a(9a + 3) is divisible by 8. Remainder 1
Case 2: (6a - 1)^2 = 36a^2 - 12a + 1 = 4a(9a - 3) + 1
Same concept as above applies here too
