Sum of Four-Digit Numbers

Let $S = a + b + c + d$. The sum of all 4-digit numbers formed by 4 distinct non-zero digits is given by: $\text{Sum} = 3! \times (a + b + c + d) \times 1111 = 6666S$

Given $6666S = 153310 + n$. Since $n$ is a single digit natural number ($1 \le n \le 9$): $153310 + n$ must be a multiple of 6666. Dividing 153310 by 6666 gives $153310 = 6666 \times 22 + 6658$. Next multiple = $6666 \times 23 = 153318$. Thus, $153310 + n = 153318 \implies n = 8$ and $S = 23$.

Value of $(a+b+c+d+n) = 23 + 8 = 31$.