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

The correct option is 31.

First, let the four distinct non-zero digits be $a$, $b$, $c$, and $d$. We want to find the sum of all four-digit numbers formed by permuting these four distinct digits. Since there are 4 distinct digits, the total number of permutations (and thus, the total number of such four-digit numbers) is:
$4!=24$

In these 24 numbers, each digit appears exactly once in each position (thousands, hundreds, tens, and units) an equal number of times. Specifically, each digit will appear in each place value exactly:
$\frac{24}{4}=6\text{times}$

Therefore, the sum of the digits in any place value (thousands, hundreds, tens, or units) across all 24 numbers is:
$6(a+b+c+d)$

By considering the place values (1000 for thousands, 100 for hundreds, 10 for tens, and 1 for units), the total sum of all these numbers is:
$\text{Sum}=1000\left[6\right(a+b+c+d\left)\right]+100\left[6\right(a+b+c+d\left)\right]+10\left[6\right(a+b+c+d\left)\right]+1\left[6\right(a+b+c+d\left)\right]$

Factoring out $6(a+b+c+d)$, we get:
$\text{Sum}=6(a+b+c+d)(1000+100+10+1)$
$\text{Sum}=6(a+b+c+d)\left(1111\right)$
$\text{Sum}=6666(a+b+c+d)$

We are given that this sum is equal to $153310+n$, where $n$ is a single-digit natural number (i.e., $n\in \{1,2,3,4,5,6,7,8,9\}$). So:
$6666(a+b+c+d)=153310+n$

To find the value of the sum $S=a+b+c+d$, we divide both sides by 6666:
$a+b+c+d=\frac{153310+n}{6666}$

Since $a,b,c,d$ are digits, their sum must be an integer. Thus, $153310+n$ must be divisible by 6666. Let's perform division on 153310 by 6666 to find the remainder:
$153310÷6666\approx 23.0003$

Multiplying 6666 by 23, we get:
$6666\times 23=153318$

Comparing this with $153310+n$:
$153310+n=153318$
$n=8$

Since $n=8$ is indeed a single-digit natural number, this is the correct value. Now, using this value, we can find the sum of the digits:
$a+b+c+d=\frac{153318}{6666}=23$

Finally, we need to calculate the value of $\mathrm{(a}+b+c+d+\mathrm{n)}$:
$(a+b+c+d+n)=23+8=31$