If a perfect square,not divisible by 6,be div | vedclass

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(A) Let the number be $n^2$. Since $n^2$ is not divisible by $6$,$n$ is not divisible by $2$ or $3$.Any integer $n$ can be expressed in the form $6k \

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$1, 3$ or $5$

Vedclass · Answer

(A) Let the number be $n^2$. Since $n^2$ is not divisible by $6$,$n$ is not divisible by $2$ or $3$.Any integer $n$ can be expressed in the form $6k \pm 1$ or $6k \pm 2$ or $6k \pm 3$.If $n$ is not divisible by $2$ or $3$,$n$ must be of the form $6k \pm 1$.Squaring this,we get $n^2 = (6k \pm 1)^2 = 36k^2 \pm 12k + 1$.$n^2 = 6(6k^2 \pm 2k) + 1$.Thus,the remainder when $n^2$ is divided by $6$ is $1$.However,checking other squares not divisible by $6$:$1^2 = 1$ (remainder $1$)$5^2 = 25 = 6 	imes 4 + 1$ (remainder $1$)$7^2 = 49 = 6 	imes 8 + 1$ (remainder $1$)$11^2 = 121 = 6 	imes 20 + 1$ (remainder $1$)Since all such squares leave a remainder of $1$,and the options provided are sets,the most appropriate choice based on standard modular arithmetic properties for squares not divisible by $2$ or $3$ is $1$.

If a perfect square,not divisible by $6$,be divided by $6$,the remainder will be

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