To determine the mean of all 4-digit even natural numbers represented as 'aabb', where a is greater than 0, execute the following steps:
- Represent the number 'aabb' algebraically: 1000a + 100a + 10b + b = 1100a + 11b.
- For the number to be even, the units digit, b, must be an even number. Therefore, b can assume values from the set {0, 2, 4, 6, 8}.
- Given that 'aabb' is a 4-digit number and a>0, the possible values for a range from 1 to 9.
- Compute the aggregate of all permissible numbers:
Total sum = Σ(1100a + 11b)
The summation is performed where a iterates from 1 to 9 and b takes values from {0, 2, 4, 6, 8}.
Total sum = 1100Σa + 11Σb.
| Component | Calculation of Sum | Summation Value |
|---|
| Sum of possible values for a | Σa for a = 1 through 9 | 45 |
| Sum of possible values for b | Σb for b ∈ {0, 2, 4, 6, 8} | 20 |
| Count of distinct values for a | 9 | 9 |
| Count of distinct values for b | 5 | 5 |
Contribution from a: 1100 × 45 = 49500.
Contribution from b: 11 × 20 = 220.
The overall sum is calculated as: (1100 × Σa) × (count of b) + (11 × Σb) × (count of a) = (49500 × 5) + (220 × 9) = 247500 + 1980 = 249480.
The total count of such numbers is 9 × 5 = 45.
The mean is derived by dividing the total sum by the total count: 249480 ÷ 45 = 5544.
Consequently, the mean of all 4-digit even natural numbers of the form 'aabb' is calculated to be 5544.