\(23_{five} = X_{ten}; X_{ten}\) = 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\(_2\)