The sum of all four-digit numbers that can be formed with the distinct non-zero digits $a, b, c,$ and $d$, with each digit appearing exactly once in every number, is $153310 + n$, where $n$ is a single digit natural number. Then, the value of $(a + b + c + d + n)$ is

We need to find the sum of all four-digit numbers formed using four distinct non-zero digits a, b, c, and d, where each digit appears exactly once in every number.

---

Total number of 4-digit numbers possible = $4! = 24$ numbers

We have 4 choices for the first position, 3 remaining choices for the second position, 2 for the third, and 1 for the last position. So: $4 \times 3 \times 2 \times 1 = 24$.

---

Since we're forming ALL possible arrangements, each digit will appear an equal number of times in each position.

Each digit appears in each position exactly $24 \div 4 = 6$ times.

Out of 24 numbers, digit 'a' will be in the thousands place 6 times, hundreds place 6 times, tens place 6 times, and units place 6 times. Same for b, c, and d.

---

For any digit x, its total contribution across all 24 numbers is:

6 times in thousands place: $6 \times x \times 1000 = 6000x$

6 times in hundreds place: $6 \times x \times 100 = 600x$

6 times in tens place: $6 \times x \times 10 = 60x$

6 times in units place: $6 \times x \times 1 = 6x$

Total contribution of digit x = $6000x + 600x + 60x + 6x = 6666x$

---

Sum of all numbers = $6666a + 6666b + 6666c + 6666d = 6666(a + b + c + d)$

---

Given: $6666(a + b + c + d) = 153310 + n$

To find $(a + b + c + d)$, we divide: $153310 \div 6666$

$153310 \div 6666 = 23.001...$

Since $(a + b + c + d)$ must be a whole number (sum of digits), we take the integer part: $a + b + c + d = 23$

---

If $a + b + c + d = 23$, then:

$6666 \times 23 = 153318$

Comparing with our equation:

$6666 \times 23 = 153310 + n$

$153318 = 153310 + n$

$n = 8$

---

Therefore, $a + b + c + d + n = 23 + 8 = 31$

$n = 8$ is indeed a single-digit natural number as required, and our calculation checks out perfectly.