Evaluate \( 202^2_{three} - 112^2_{three}\)
21120
21121
21112
21011
Correct answer is A
\(202^2_{three}\)when converted to base ten \(=(202_3)^2\\
202_3 = 2 \times 3^2 + 0 \times 3^1 + 2\times 3^0 = 18 + 0 + 2\\
=20_{ten}; (202_3)^2 = (20)^2_{ten} = 400\\
112^2_{three}\)when converted to base ten \(= (112_3)^2\\
112_3 = 1 \times 3^2 + 1 \times 3^1 + 2\times 3^0 = 9+3+2=14_{ten}\\
(112_3)^2 = (14)^2_{ten} = 196_{ten}\\
Evaluate \Longrightarrow 400-196 = 204\)
Reconvert to base three
\(\begin{matrix}
3 & 204\\
3 & 69 &R0\\
3 & 22 & R2\\
3 & 7 & R1\\
2 & 2 & R1\\
& 0& R2 \uparrow\\
\end{matrix} \\
=21120_3\)