To solve the problem step by step, we need to find the least number \( x \) that meets the specified conditions:
1. **Understanding the Problem**:
We want to find the least four-digit number \( x \) such that when \( x \) is divided by 2, 3, 4, 5, 6, and 7, it leaves a remainder of 1. This means that \( x - 1 \) must be divisible by all these numbers.
2. **Finding the LCM**:
To find \( x - 1 \), we need to calculate the least common multiple (LCM) of the numbers 2, 3, 4, 5, 6, and 7.
- The prime factorization of each number is:
- \( 2 = 2^1 \)
- \( 3 = 3^1 \)
- \( 4 = 2^2 \)
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)
- \( 7 = 7^1 \)
- The LCM is found by taking the highest power of each prime:
- \( 2^2 \) from 4,
- \( 3^1 \) from 3 or 6,
- \( 5^1 \) from 5,
- \( 7^1 \) from 7.
- Thus, the LCM is:
\[
\text{LCM} = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7
\]
- Calculating this step by step:
- \( 4 \times 3 = 12 \)
- \( 12 \times 5 = 60 \)
- \( 60 \times 7 = 420 \)
3. **Finding \( x \)**:
Since \( x - 1 \) must be a multiple of 420, we can express \( x \) as:
\[
x = 420k + 1
\]
where \( k \) is a positive integer.
4. **Finding the Range**:
We need \( x \) to be a four-digit number between 2000 and 2500:
\[
2000 \leq 420k + 1 \leq 2500
\]
Subtracting 1 from all parts gives:
\[
1999 \leq 420k \leq 2499
\]
Dividing by 420:
\[
\frac{1999}{420} \leq k \leq \frac{2499}{420}
\]
Calculating the bounds:
- \( \frac{1999}{420} \approx 4.76 \) (so \( k \geq 5 \))
- \( \frac{2499}{420} \approx 5.95 \) (so \( k \leq 5 \))
Therefore, the only integer value for \( k \) is 5.
5. **Calculating \( x \)**:
Substituting \( k = 5 \):
\[
x = 420 \times 5 + 1 = 2100 + 1 = 2101
\]
6. **Finding the Sum of the Digits**:
The digits of 2101 are 2, 1, 0, and 1. Thus, the sum of the digits is:
\[
2 + 1 + 0 + 1 = 4
\]
**Final Answer**:
The sum of the digits of \( x \) is \( \boxed{4} \).
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