The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7 is ____________.

The correct answer is 569.

We are asked to find the total number of 4-digit integers in the closed interval [2022, 4482] using the set of digits S = {0, 2, 3, 4, 6, 7}. Repetition of digits is allowed unless restricted by place value analysis.

Let the 4-digit number be represented as d₁d₂d₃d₄, where d₁, d₂, d₃, d₄ ∈ {0, 2, 3, 4, 6, 7}.

We break the problem into cases based on the first digit d₁:

Case 1: Numbers starting with 2 (d₁ = 2)

The number is of the form 2d₂d₃d₄ and must be ≥ 2022.

• Subcase 1a: d₂ > 0 (i.e., d₂ ∈ {2, 3, 4, 6, 7} - 5 options)

For each choice of d₂, d₃ can be any of the 6 digits, and d₄ can be any of the 6 digits.

Number of possibilities = 5 × 6 × 6 = 180.

• Subcase 1b: d₂ = 0 (the number is 20d₃d₄)

We need 20d₃d₄ ≥ 2022.

- If d₃ > 2 (i.e., d₃ ∈ {3, 4, 6, 7} - 4 options): d₄ can be any of the 6 digits.

Number of possibilities = 4 × 6 = 24.

- If d₃ = 2 (i.e., 202d₄): d₄ must be ≥ 2 (i.e., d₄ ∈ {2, 3, 4, 6, 7} - 5 options).

Number of possibilities = 5.

- If d₃ = 0 (i.e., 200d₄): This gives numbers < 2022, so 0 possibilities.

Total for Case 1 = 180 + 24 + 5 = 209.

Case 2: Numbers starting with 3 (d₁ = 3)

The number is of the form 3d₂d₃d₄.

Since 3000 ≤ 3d₂d₃d₄ ≤ 3777, all such numbers lie strictly within [2022, 4482].

d₂ can be any of the 6 digits.

d₃ can be any of the 6 digits.

d₄ can be any of the 6 digits.

Total for Case 2 = 1 × 6 × 6 × 6 = 216.

Case 3: Numbers starting with 4 (d₁ = 4)

The number is of the form 4d₂d₃d₄ and must be ≤ 4482.

• Subcase 3a: d₂ ∈ {0, 2, 3} (3 options)

For each choice of d₂, d₃ can be any of the 6 digits, and d₄ can be any of the 6 digits.

Number of possibilities = 3 × 6 × 6 = 108.

• Subcase 3b: d₂ = 4 (the number is 44d₃d₄)

We need 44d₃d₄ ≤ 4482.

- If d₃ ∈ {0, 2, 3, 4, 6, 7} (all digits from S are ≤ 7): Since the maximum digit available is 7, for any d₃ ∈ S, 44d₃d₄ ≤ 4477 ≤ 4482.

Thus, all 6 choices for d₃ and all 6 choices for d₄ are valid.

Number of possibilities = 6 × 6 = 36.

• Subcase 3c: d₂ ∈ {6, 7}: These give numbers ≥ 4600 > 4482, so 0 possibilities.

Total for Case 3 = 108 + 36 = 144.

Total Calculation:

Summing up the counts from all valid cases:

Total numbers = (Case 1) + (Case 2) + (Case 3)

$\text{Total}=209+216+144=569$

Therefore, the total number of 4-digit integers in the interval [2022, 4482] formed using the given digits is 569.