There is a remainder of 3 when a number is divided by 6. What will be the remainder if the square of each number will be 6?
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Let the number be 'N'
N/6 == Quotient + Remainder
N= 6* Quotient + 3
N^2 = 6^2*Quotient^2 + 9 + 2*3*6*Quotient
N^2 = 36*Quotient^2 + 9 + 36*Quotient
Now,
N^2/6 = 36*Quotient^2/6 + 9/6 + 36*Quotient/6
Now the remainder is nothing but the remainder of 9/6 [since all other terms are perfectly divisible by '6']
Hence remainder is '3'
Hope it helps.
N/6 == Quotient + Remainder
N= 6* Quotient + 3
N^2 = 6^2*Quotient^2 + 9 + 2*3*6*Quotient
N^2 = 36*Quotient^2 + 9 + 36*Quotient
Now,
N^2/6 = 36*Quotient^2/6 + 9/6 + 36*Quotient/6
Now the remainder is nothing but the remainder of 9/6 [since all other terms are perfectly divisible by '6']
Hence remainder is '3'
Hope it helps.
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