1. **Problem (a):** Given that $231_x - 143_x = 44_x$, find the value of $x$. 2. **Understanding the problem:** The numbers are in base $x$. We need to convert each number from base $x$ to base 10, perform the subtraction, and then equate it to the base 10 value of $44_x$. 3. **Convert each number to base 10:** - $231_x = 2x^2 + 3x + 1$ - $143_x = 1x^2 + 4x + 3$ - $44_x = 4x + 4$ 4. **Set up the equation:** $$ (2x^2 + 3x + 1) - (1x^2 + 4x + 3) = 4x + 4 $$ 5. **Simplify the left side:** $$ 2x^2 + 3x + 1 - 1x^2 - 4x - 3 = 4x + 4 $$ $$ (2x^2 - 1x^2) + (3x - 4x) + (1 - 3) = 4x + 4 $$ $$ x^2 - x - 2 = 4x + 4 $$ 6. **Bring all terms to one side:** $$ x^2 - x - 2 - 4x - 4 = 0 $$ $$ x^2 - 5x - 6 = 0 $$ 7. **Solve the quadratic equation:** $$ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-6)}}{2} = \frac{5 \pm \sqrt{25 + 24}}{2} = \frac{5 \pm \sqrt{49}}{2} $$ $$ x = \frac{5 \pm 7}{2} $$ 8. **Possible solutions:** - $x = \frac{5 + 7}{2} = 6$ - $x = \frac{5 - 7}{2} = -1$ (not valid since base must be positive integer greater than digits) 9. **Check base validity:** The digits in the numbers are up to 4, so base $x$ must be greater than 4. $x=6$ is valid. **Final answer for (a):** $x = 6$ --- 1. **Problem (b):** If $y$ is a positive integer, calculate the least value of $y$ for which $$ 2y + 13 \equiv 3 \pmod{9} $$ 2. **Rewrite the congruence:** $$ 2y + 13 \equiv 3 \pmod{9} $$ $$ 2y \equiv 3 - 13 \pmod{9} $$ $$ 2y \equiv -10 \pmod{9} $$ 3. **Simplify modulo 9:** $$ -10 \equiv -10 + 18 = 8 \pmod{9} $$ So, $$ 2y \equiv 8 \pmod{9} $$ 4. **Find the multiplicative inverse of 2 modulo 9:** Since $2 \times 5 = 10 \equiv 1 \pmod{9}$, the inverse of 2 mod 9 is 5. 5. **Multiply both sides by 5:** $$ y \equiv 8 \times 5 = 40 \equiv 40 - 36 = 4 \pmod{9} $$ 6. **Least positive integer solution:** $$ y = 4 $$ **Final answer for (b):** $y = 4$