Finding the Area Defined by Given Inequalities

Given inequalities:

• \( y \geq x + 4 \)
• \( -4 \leq x^2 + y^2 + 4(x – y) \leq 0 \)

Step 1: Simplify the second inequality

\[ x^2 + y^2 + 4x – 4y = (x + 2)^2 – 4 + (y – 2)^2 – 4 = (x + 2)^2 + (y – 2)^2 – 8 \] So the inequality becomes: \[ -4 \leq (x + 2)^2 + (y – 2)^2 – 8 \leq 0 \Rightarrow 4 \leq (x + 2)^2 + (y – 2)^2 \leq 8 \]

This represents an annular region (ring) centered at \( (-2, 2) \), with:

• Inner radius: \( \sqrt{4} = 2 \)
• Outer radius: \( \sqrt{8} = 2\sqrt{2} \)

Step 2: Apply the line constraint \( y \geq x + 4 \)

The line cuts the annular region approximately in half (passing near the center), so the required area is half of the total area of the ring.

Step 3: Calculate Area

Area of full ring: \[ \pi(R^2 – r^2) = \pi((2\sqrt{2})^2 – 2^2) = \pi(8 – 4) = 4\pi \] Area of the required region: \[ \frac{4\pi}{2} = \boxed{2\pi} \]

✅ Final Answer: \( \boxed{2\pi} \)

Finding the Value of \( a + b + c + d + n \)

Given: The sum of all four-digit numbers formed using the digits \( a, b, c, d \) exactly once in each number is:

\[ 153310 + n \] where \( n \) is a single-digit natural number.

Step 1: Total Permutations

Total 4-digit numbers formed using all four digits once = \( 4! = 24 \)

Step 2: Positional Contribution

Each digit appears \( \frac{4!}{4} = 6 \) times in each place (units, tens, hundreds, thousands).
Let \( S = a + b + c + d \). Then total sum is: \[ \text{Total Sum} = 6 \times S \times (1000 + 100 + 10 + 1) = 6 \times S \times 1111 \] So, \[ 6 \times S \times 1111 = 153310 + n \tag{1} \]

Step 3: Try Values for \( S \)

Try \( S = 23 \Rightarrow 6 \cdot 23 \cdot 1111 = 153318 \Rightarrow n = 8 \)
✅ This satisfies the equation.

Step 4: Final Calculation

\[ a + b + c + d + n = 23 + 8 = \boxed{31} \]

✅ Final Answer: \( \boxed{31} \)