When a^2 is divided by 6,which of the followi | vedclass

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(B) Any natural number $a$ can be expressed in the form $6k, 6k+1, 6k+2, 6k+3, 6k+4,$ or $6k+5$. Calculating $a^2 \pmod{6}$ for each case: $1) (6k)^2

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$2$

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(B) Any natural number $a$ can be expressed in the form $6k, 6k+1, 6k+2, 6k+3, 6k+4,$ or $6k+5$. Calculating $a^2 \pmod{6}$ for each case: $1) (6k)^2 = 36k^2 = 6(6k^2) \equiv 0 \pmod{6}$ $2) (6k+1)^2 = 36k^2 + 12k + 1 = 6(6k^2 + 2k) + 1 \equiv 1 \pmod{6}$ $3) (6k+2)^2 = 36k^2 + 24k + 4 = 6(6k^2 + 4k) + 4 \equiv 4 \pmod{6}$ $4) (6k+3)^2 = 36k^2 + 36k + 9 = 6(6k^2 + 6k + 1) + 3 \equiv 3 \pmod{6}$ $5) (6k+4)^2 = 36k^2 + 48k + 16 = 6(6k^2 + 8k + 2) + 4 \equiv 4 \pmod{6}$ $6) (6k+5)^2 = 36k^2 + 60k + 25 = 6(6k^2 + 10k + 4) + 1 \equiv 1 \pmod{6}$ The possible remainders are $0, 1, 3, 4$. Thus,the remainders $2$ and $5$ are not possible.

When $a^2$ is divided by $6$,which of the following cannot be the remainder? $(a \in N)$

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