On dividing a number by 5,we get 3 as the rem | vedclass

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(D) Let the number be $n$. According to the problem,when $n$ is divided by $5$,the remainder is $3$. This can be expressed as $n = 5k + 3$,where $k$ i

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$4$

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(D) Let the number be $n$. According to the problem,when $n$ is divided by $5$,the remainder is $3$. This can be expressed as $n = 5k + 3$,where $k$ is an integer.We need to find the remainder when $n^2$ is divided by $5$.$n^2 = (5k + 3)^2$$n^2 = 25k^2 + 30k + 9$$n^2 = 25k^2 + 30k + 5 + 4$$n^2 = 5(5k^2 + 6k + 1) + 4$Since $5(5k^2 + 6k + 1)$ is divisible by $5$,the remainder when $n^2$ is divided by $5$ is $4$.

On dividing a number by $5$,we get $3$ as the remainder. What will be the remainder when the square of this number is divided by $5$?

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