1110
10111
11101
111100
Correct answer is B
\(23_{five} = X_{ten}; X_{ten} = 2\times 5^1 + 3\times 5^0 = 10 + 3 = 13\\
101_{three}=P_{ten}; P_{ten} = 1\times 3^2 + 0\times 3^1 + 1\times 3^0=9+0+1=10_{ten}\\
Y = 13+10=23_{ten}\);
Converting to base two
\(\begin{matrix}
2 & 23\\
2 & 11 &R1\\
2 & 5 & R1\\
2 & 2 & R1\\
2 & 1 & R0\\
& 0& R1 \uparrow\\
\end{matrix} \\
=y=10111_2\)
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