Question

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 $(a + b + c + d)$ is ?

Answer 1

There are 24 distinct four-digit numbers that can be formed with the digits $a$, $b$, $c$, and $d$ (since there are $4! = 24$ possible permutations). The sum of all these numbers is:

$24 \times (a + b + c + d) \times 1111$.

We are given that this sum is $153310 + n$, where $n$ is a single digit. By equating, we have:

$24 \times (a + b + c + d) \times 1111 = 153310 + n$.

From this equation, solve for $a + b + c + d + n$.

Answer 2

There are 24 distinct four-digit numbers that can be formed with the digits $a$, $b$, $c$, and $d$ (since there are $4! = 24$ possible permutations). The sum of all these numbers is:

$24 \times (a + b + c + d) \times 1111$.

We are given that this sum is $153310 + n$, where $n$ is a single digit. By equating, we have:

$24 \times (a + b + c + d) \times 1111 = 153310 + n$.

From this equation, solve for $a + b + c + d + n$.