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:

Question

Accepted Answer

The problem involves calculating the value of $$(a + b + c + d + n)$$, where $$a, b, c, d$$ are distinct non-zero digits used to form all possible four-digit numbers. Here's the step-by-step solution:

**Total Numbers Formed:**
   - With four distinct digits $$a, b, c, d$$, the total number of four-digit numbers is:
     $$
     4! = 24
     $$

**Sum of All Numbers:**
   - Each digit appears in every place value (units, tens, hundreds, and thousands) exactly $$6$$ times (since $$3! = 6$$).
   - The contribution of each digit to the sum is:
     $$
     (a + b + c + d) 	imes 6 	imes (1 + 10 + 100 + 1000)
     $$
   - Simplify:
     $$
     	ext{Sum of all numbers} = (a + b + c + d) 	imes 6 	imes 1111 = 6666 	imes (a + b + c + d)
     $$

**Given Information:**
   - The sum of all numbers is given as $$153310 + n$$, where $$n$$ is a single-digit natural number.
   - Dividing $$153310$$ by $$1111$$, the remainder is:
     $$
     153310 \mod 1111 = 1103
     $$
   - To make the sum divisible by $$1111$$, add $$n = 8$$:
     $$
     153310 + n = 153318
     $$

**Determine $$(a + b + c + d)$$:**
   - The total sum is now divisible by $$1111$$:
     $$
     \frac{153318}{1111} = 138
     $$
   - From the equation:
     $$
     (a + b + c + d) 	imes 6 = 138
     $$
   - Solve for $$(a + b + c + d)$$:
     $$
     a + b + c + d = \frac{138}{6} = 23
     $$

**Final Calculation:**
   - The required value is:
     $$
     a + b + c + d + n = 23 + 8 = 31
     $$

### Final Answer:

The value of $$(a+b+c+d+n)$$ is **31**.  
Correct answer: **31**

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