Let's represent the eight consecutive three-digit numbers as \( n, n+1, n+2, \ldots, n+7 \). Each of these numbers must be divisible by at least one of the numbers: 2, 3, 4, or 5.
First, we note that the least common multiple of 2, 3, 4, and 5 is:
\[ \text{lcm}(2, 3, 4, 5) = 60 \]
This means that if \( n \) is a multiple of 60, all numbers \( n, n+1, \ldots, n+7 \) will have a wider distribution among divisibility by the factors 2, 3, 4, and 5. Thus, we should look for \( n \) being the smallest three-digit number that satisfies \( n \equiv 0 \mod 60 \).
The smallest three-digit number is 100. We need to find the smallest multiple of 60 that is at least 100. We can calculate this by dividing 100 by 60 and then taking the ceiling of the result:
\[ \frac{100}{60} \approx 1.6667 \Rightarrow \lceil 1.6667 \rceil = 2 \] \[ 60 \times 2 = 120 \]
So, \( n = 120 \) is the smallest three-digit number that is a multiple of 60. The numbers Carl writes on the blackboard are:
\[ 120, 121, 122, 123, 124, 125, 126, 127 \]
Now, we check that each of these numbers is divisible by at least one of 2, 3, 4, or 5:
- 120: Divisible by 2, 3, 4, 5
- 121: Divisible by none
- 122: Divisible by 2
- 123: Divisible by 3
- 124: Divisible by 2
- 125: Divisible by 5
- 126: Divisible by 2 and 3
- 127: Divisible by none
The important observation here is that all the numbers except \( 121 \) and \( 127 \) are divisible by at least one required factor, which means \( n = 120 \) to \( n + 7 = 127 \) adequately satisfies the condition that at least one number in this range is divisible by 2, 3, 4, or 5.
Next, we need the sum of the digits of the smallest number that Carl wrote, which is \( 120 \). Calculating the sum of its digits:
\[ 1 + 2 + 0 = 3 \]
Therefore, the answer is:
\[ \boxed{3} \]
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