1.

The remainder of any perfect square divided by 3 is (A) 0 (B) 1 (C) either (A) or (B) (D) neither (A) nor (B)

Answer»

Correct option is (C) either (A) or (B)

Let \(n^2\) is a perfect square.

Then n is a natural number.

\(\because\) n is any arbitrary natural number then for any integers a & b we have n = 3a+b, where \(0\leq b<3.\)  (By division algorithm)

\(\therefore\) Possible values for b is 0, 1 and 2.

Therefore, any natural number is of the form 3m, 3m+1 and 3m+2, where \(m\in N.\)

Case I :- If n = 3m

Then \(n^2=3m^2\)

Thus, \(31n^2\)

\(\therefore\) Remainder of \(n^2\) when divided by 3 is 0.

Case II :- If n = 3m+1

Then \(n^2=(3m+1)^2=9m^2+6m+1\) \(=3(3m^2+2m)+1.\)

Thus, when \(n^2\) is divided by 3 then leaving remainder is 1.

Case III :- If n = 3m+2

Then \(n^2=(3m+2)^2=9m^2+12m+4\) \(=3(3m^2+4m+1)+1.\)

Thus, when \(n^2\) is divided by 3 then leaving remainder is 1.

Hence for all possibilities of n, when \(n^2\) is divided by 3 then leaving remainder is either 0 or 1.

Correct option is (C) either (A) or (B)



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