To find the sum \(a + b + c + d\), given that the sum of all four-digit numbers formed by the distinct digits \(a\), \(b\), \(c\), and \(d\) is \(153310 + n\), we first analyze the problem mathematically.
All possible four-digit numbers are formed using permutations of these four distinct digits. There are \(4!\) (which equals 24) such permutations.
Each digit appears equally often in each position (thousands, hundreds, tens, and units). Therefore, each digit appears \(24 \div 4 = 6\) times in each position.
The value contributed by a digit \(x\) in different positions is:
\( \begin{aligned} &\text{Thousands place: } 1000 \times x, \\ &\text{Hundreds place: } 100 \times x, \\ &\text{Tens place: } 10 \times x, \\ &\text{Units place: } x. \end{aligned} \)
The sum of positional values is \(1000+100+10+1 = 1111\).
Since each digit appears 6 times, the total contribution of a single digit across all positions is \(6 \times 1111 \times \text{the digit}\).
The total sum of all numbers formed is therefore \(6 \times 1111 \times (a+b+c+d) = 6666(a+b+c+d).\)
We are given that the sum of all numbers is \(153310 + n\). Equating this to our derived sum, and assuming \(n=4\) (to make it a single digit), we have:
\(6666 \times (a+b+c+d) = 153310 + 4.\)
This simplifies to:
\(6666 \times (a+b+c+d) = 153314.\)
Solving for \((a+b+c+d)\):
\(a+b+c+d = \frac{153314}{6666} = 23.\)
This calculation confirms that \(a+b+c+d = 23\).
The value of the sum \((a+b+c+d)\) is:
23