1. **Problem (a):** Convert a number $abc.de$ in base $x$ to base 10 using the expansion method. The formula is: $$abc.de_x = a \times x^2 + b \times x^1 + c \times x^0 + d \times x^{-1} + e \times x^{-2}$$ where $a,b,c,d,e$ are digits in base $x$. This means each digit is multiplied by the base raised to the power of its position, counting left of the decimal as positive powers and right as negative powers. 2. **Problem (b)(i):** Convert binary number $10.0011_2$ to decimal. Write the place values: $$1 \times 2^1 + 0 \times 2^0 + 0 \times 2^{-1} + 0 \times 2^{-2} + 1 \times 2^{-3} + 1 \times 2^{-4}$$ Calculate each term: $$= 1 \times 2 + 0 + 0 + 0 + 1 \times \frac{1}{8} + 1 \times \frac{1}{16} = 2 + 0 + 0 + 0 + 0.125 + 0.0625 = 2.1875$$ 3. **Problem (b)(ii):** Find place values of digits 1 in $(0.0011)_2$. Digits are at positions $-3$ and $-4$: $$1 \times 2^{-3} = \frac{1}{8} = 0.125$$ $$1 \times 2^{-4} = \frac{1}{16} = 0.0625$$ 4. **Problem (b)(iii):** Sum $(1+1+1+1)$ in binary. Sum in decimal is $4$. Convert $4$ to binary: $$4_{10} = 100_2$$ 5. **Problem (b)(iv):** Calculate $101_2 \div 10_2$ using long division in base 2. $101_2 = 5_{10}$, $10_2 = 2_{10}$. Divide 5 by 2: $2 \times 2 = 4$, remainder $1$. So quotient is $10_2$ (which is 2 in decimal), remainder $1$. 6. **Problem (c):** Find $x$ in base 2 from equation: $$11_x = 1000_x + 101_x$$ Convert each to decimal: $$11_x = 1 \times x + 1 = x + 1$$ $$1000_x = 1 \times x^3 = x^3$$ $$101_x = 1 \times x^2 + 0 + 1 = x^2 + 1$$ Equation: $$x + 1 = x^3 + x^2 + 1$$ Simplify: $$x + 1 - 1 = x^3 + x^2$$ $$x = x^3 + x^2$$ Divide both sides by $x$ (assuming $x \neq 0$): $$\cancel{x} = x^2 + x \Rightarrow \cancel{x} = x^2 + x$$ $$1 = x + 1$$ Subtract 1: $$1 - 1 = x + 1 - 1$$ $$0 = x$$ Since $x=0$ is invalid base, check original equation again. Rewrite: $$x = x^3 + x^2$$ $$0 = x^3 + x^2 - x$$ $$0 = x(x^2 + x - 1)$$ So either $x=0$ (invalid) or solve: $$x^2 + x - 1 = 0$$ Use quadratic formula: $$x = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$$ Since base must be integer $>1$, no integer solution. Therefore, no valid base $x$ satisfies this equation. 7. **Problem (d):** Given in base $b$: $$AB_b + BA_b = 121_{10}$$ Where $A \neq B$ are digits. Express in decimal: $$AB_b = A \times b + B$$ $$BA_b = B \times b + A$$ Sum: $$AB_b + BA_b = (A b + B) + (B b + A) = (A + B) b + (A + B) = (A + B)(b + 1)$$ Given: $$(A + B)(b + 1) = 121$$ Factor 121: $$121 = 11 \times 11$$ Possible pairs for $(A+B)$ and $(b+1)$ are $(11,11)$ or $(1,121)$ or $(121,1)$. Since $A$ and $B$ are digits, $A+B \leq 2(b-1)$, and $b$ must be at least max digit + 1. Try $(A+B) = 11$, $(b+1) = 11 \Rightarrow b=10$. Digits $A$ and $B$ sum to 11, digits less than base 10. Try $A=5$, $B=6$ (or vice versa), $A \neq B$. Check: $$(5+6)(10+1) = 11 \times 11 = 121$$ Valid. **Final answers:** (a) $$abc.de_x = a x^2 + b x + c + d x^{-1} + e x^{-2}$$ (b)(i) $$10.0011_2 = 2.1875_{10}$$ (b)(ii) Place values of 1's in $(0.0011)_2$ are $2^{-3} = 0.125$ and $2^{-4} = 0.0625$ (b)(iii) Sum $(1+1+1+1)_2 = 100_2$ (b)(iv) $101_2 \div 10_2 = 10_2$ remainder $1_2$ (c) No valid integer base $x$ satisfies $11_x = 1000_x + 101_x$ (d) Base $b=10$, digits $A=5$, $B=6$ (or $A=6$, $B=5$) satisfy $AB_b + BA_b = 121_{10}$