The sum of all four-digit numbers that can be formed with the dist inct 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

When gives n distinct numbers, and asked to find the sum of the possible n distinct numbers that can be formed, 
We use the formula, $$\left(10^{n-1}+10^{n-2}+10^{n-3}+...\right)\left(\left(n-1\right)!\right)\left(Sum\ of\ numbers\right)$$

Inserting n=4, $$\left(1000+100+10+1\right)\left(3!\right)\left(a+b+c+d\right)$$
$$\left(6666\right)\left(a+b+c+d\right)$$
We are told that this value equals, $$153310+n$$
Since we are told that n is a single digit natural number, the total value cannot be that much greater and 6666 should perfectly divide $$153310+n$$

Upon dividing 153310 by 6666 we get the quotient as 22.99
Nearest value being 23, We take 6666x23 giving us the value 153318

Hence the value of n=8 and the value of a+b+c+d=23

Value of a+b+c+d+n=31.