Convert/Calculate as per the instruction: i) (1503) 8 = (?) 16 ii) (101000101) 2 = (?) 8 iii) (1010 + 1101) 2 (110) 2 iv) (100111) 2 + (110) 2

i) Convert $(1503{)}_8$ to $(?{)}_{16}$

Step 1: Convert $(1503{)}_8$ to decimal (base 10). $$(1503{)}_8=1\times 8^3+5\times 8^2+0\times 8^1+3\times 8^0$$ $$=1\times 512+5\times 64+0\times 8+3\times 1$$ $$=512+320+0+3=835_{10}$$

Step 2: Convert $835_{10}$ to hexadecimal (base 16). Divide by 16 and find remainders: $$835÷16=52\text{ remainder }3$$ $$52÷16=3\text{ remainder }4$$ $$3÷16=0\text{ remainder }3$$ Reading remainders from bottom to top: $(343{)}_{16}$.

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ii) Convert $(101000101{)}_2$ to $(?{)}_8$

Step 1: Group binary digits into sets of three from the right (pad with leading zeros if needed). $$(101000101{)}_2=001010001001$$

Step 2: Convert each group to octal: $$001_2=1_8,010_2=2_8,001_2=1_8,001_2=1_8$$

Final result: $(1211{)}_8$.

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iii) $(1010+1101{)}_2-(110{)}_2$

Step 1: Perform addition $(1010{)}_2+(1101{)}_2$: \begin{array}{c@{}c@{}c@{}c@{}c} & 1 & 0 & 1 & 0 \\ + & 1 & 1 & 0 & 1 \\ \hline 1 & 0 & 1 & 1 & 1 \\ \end{array} Result: $(10111{)}_2$.

Step 2: Subtract $(110{)}_2$ from $(10111{)}_2$: \begin{array}{c@{}c@{}c@{}c@{}c} & 1 & 0 & 1 & 1 & 1 \\ - & & 1 & 1 & 0 \\ \hline & 1 & 0 & 0 & 0 & 1 \\ \end{array} Final result: $(10001{)}_2$.

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iv) $(100111{)}_2+(110{)}_2$

Step 1: Align binary numbers and add: \begin{array}{c@{}c@{}c@{}c@{}c@{}c} & 1 & 0 & 0 & 1 & 1 & 1 \\ + & & & 1 & 1 & 0 \\ \hline & 1 & 0 & 1 & 1 & 0 & 1 \\ \end{array} Final result: $(101101{)}_2$.