NCHMCT JEE 2016 Question Paper with solutions for online practice

Let, the number = x
x = Q × 6 + 3
Dividend = Quotient × divisor + remainder
Put, Q = 1, 2, 3
x = 1 × 6 + 3
x = 9
∴ ${\mathit{x}}^2$ = 81
81 / 6 = 13 $\frac{3}{6}$
x = 2 × 6 + 3 = 15
${\mathit{x}}^2$ = ${\left(15\right)}^2$
= 225
225 / 6 = 37 $\frac{3}{6}$
x = 3 × 6 + 3 = 21
${\mathit{x}}^2$ = ${\left(21\right)}^2$
= 441
441 / 6 = 73 $\frac{2}{6}$
Hence, required remainder = 3.