Let Rich's 4-digit positive integer be $a b c d$, where $d$ is the units (ones) digit, $c$ is the tens digit, $b$ is the hundreds digit, and $a$ is the thousands digit of the original number .

We begin by determining which of the 4 digits is erased.

If Rich erases the digit $a$, then the 3 -digit integer that remains is $b c d$, and the sum of the two integers is $a b c d+b c d$.

The units digit of this sum is determined by adding the units digits of $a b c d$ and $b c d$, or $d+d=2 d$. In this case, the units digit of the sum is even since $2 d$ is even.

The sum of these two integers is 6031 which has an odd units digit, and so the digit that is erased cannot be $a$.

Using a similar argument, the digit that is erased cannot be $b$ or $c$, and thus must be $d$.

Next, we determine the digits $a, b, c, d$ so that

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-100.jpg?height=155&width=220&top_left_y=730&top_left_x=1012)

The hundreds column of this sum gives that $b+a$ added to the 'carry' from the tens column has units digit 0 . That is, this sum is either 0,10 or 20 .

If the sum is 0 , then $a=0$, but looking to the thousands column we see that $a$ cannot be 0 .

If the sum is 20 , then $a+b=18$ (since any digit can be at most 9 and the carry from the tens column can be at most 2).

If $a+b=18$, then $a=9$, but looking to the thousands column we see that $a$ cannot be 9 .

Therefore, $b+a$ added to the carry from the tens column is 10, and so the carry from the hundreds column to the thousands column is 1 .

The thousands column then gives $a+1=6$ and so $a=5$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-100.jpg?height=160&width=222&top_left_y=1351&top_left_x=1014)

The hundreds column gives $b+5$ added to the carry from the tens column is 10 .

If the carry from the tens column is 0 , then $b+5=10$ and so $b=5$.

In the tens column, if $b=5$ then $c+b=c+5$ is greater than 3 and so must be 13 .

This tells us that the carry from the tens colunn cannot be 0 .

If the carry from the tens column is 1 , then $b+5+1=10$ and so $b=4$.

(Note that sum of the tens column has units digit 3 and so the carry cannot be 2.)

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-100.jpg?height=155&width=220&top_left_y=1836&top_left_x=1012)

The units digit of the sum in the tens column is 3 , and so $c+4$ added to the carry from the units column is 13 (it can't be 3 or 23).

If there is no carry from the units column, then $c+4=13$ and so $c=9$.

But if $c=9$ then the units column gives that $d+9$ must be 11 and so there is carry from the units column. Hence $c$ is not 9 .

If the carry from the units column is 1 , then $c+4+1=13$ and so $c=8$ and $d=3$.

The final sum is shown here:

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-100.jpg?height=152&width=215&top_left_y=2342&top_left_x=1020)

The sum of the digits of the original 4-digit number is $a+b+c+d=5+4+8+3=20$.
Rich chooses a 4-digit positive integer. He erases one of the digits of this integer. The remaining digits, in their original order, form a 3-digit positive integer. When Rich adds this 3 -digit integer to the original 4 -digit integer, the result is 6031 . What is the sum of the digits of the original 4-digit integer?
(A) 18
(B) 20
(C) 22
(D) 19
(E) 21