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When the square of any odd number, greater than 1, is divided by 8, it always leaves remainder :
Answer & Solution
Correct Answer:
Option
A
Let the number be N = 2x + 1
N2 = (2x + 1)2
= 4x2 + 1 + 4x
= 4x (x + 1) + 1
Clearly, 4x (x + 1) is always divisible by 8 since one of x and (x + 1) is even which when multiplied by 4 is always divisible by 8.
Hence, required remainder = 1
N2 = (2x + 1)2
= 4x2 + 1 + 4x
= 4x (x + 1) + 1
Clearly, 4x (x + 1) is always divisible by 8 since one of x and (x + 1) is even which when multiplied by 4 is always divisible by 8.
Hence, required remainder = 1
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