When (6k+1)^2 is divided by 6,the remainder i | vedclass

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(B) We are given the expression $(6k+1)^2$.Expanding this using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$,we get:$(6k+1)^2 = (6k)^2 + 2(6k)(1

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$1$

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(B) We are given the expression $(6k+1)^2$.Expanding this using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$,we get:$(6k+1)^2 = (6k)^2 + 2(6k)(1) + 1^2$$= 36k^2 + 12k + 1$We can factor out $6$ from the first two terms:$= 6(6k^2 + 2k) + 1$Let $m = 6k^2 + 2k$,where $m$ is an integer.Then the expression becomes $6m + 1$.When $6m + 1$ is divided by $6$,the quotient is $m$ and the remainder is $1$.

When $(6k+1)^2$ is divided by $6$,the remainder is . . . . . . .

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