2. If we omit one digit from a four-digit number and then add the resulting three-digit number to the original four-digit number, we get 1254. Determine the original four-digit number.
2. If we omit one digit from a four-digit number and then add the resulting three-digit number to the original four-digit number, we get 1254. Determine the original four-digit number.

## Solution.

If the thousands digit were 2 or more, the sum would be greater than 1254.

The thousands digit must be 1.

The four-digit number is of the form $\overline{1 a b c}$.

If we omit the digit $c$, we get:

```
\(\overline{1 a b c}+\overline{1 a b}=1254\)
\(1000+100 a+10 b+c+100+10 a+b=1254\)
\(110 a+11 b+c+1100=1254\)
\(110 a+11 b+c=154\)
\(a=1\) because for \(a=2\) the sum is greater than 154, and for \(a=0\) the expression \(11 b+c\) is less than 154.
\(11 b+c=44\)
\(b=4, c=0 \quad\) The number is 1140.
```

If we omit the digit $b$, we get:

```
\(\overline{1 a b c}+\overline{1 a c}=1254\)
\(1000+100 a+10 b+c+100+10 a+c=1254\)
\(110 a+10 b+2 c+1100=1254\)
\(110 a+10 b+2 c=154\)
\(a=1\)
\(10 b+2 c=44\)
\(5 b+c=22\)
\(b=3, c=7 \quad\) The number is 1137.
\(b=4, c=2 \quad\) The number is 1142.
```

If we omit the digit $a$, we get:

```
\(\overline{1 a b c}+\overline{1 b c}=1254\)
\(1000+100 a+10 b+c+100+10 b+c=1254\)
\(100 a+20 b+2 c+1100=1254\)
\(100 a+20 b+2 c=154\)
\(a=0\)
\(a=1\)
\(20 b+2 c=154\)
\(20 b+2 c=54\)
\(10 b+c=77\)
\(10 b+c=27\)
\(b=7, c=7\)
```

The number is 1077.
\(b=2, c=7\)

If we omit the digit 1, we get:

```
\(\overline{1 a b c}+\overline{a b c}=1254\)
\(200 a+20 b+2 c+1000=1254\)
\(200 a+20 b+2 c=254\)
\(100 a+10 b+c=127\)
\(a=1\)
\(10 b+c=27\)
\(b=2, c=7 \quad\) The number is 1127, but we already have this number.
```

All possible solutions are:

1077, 1127, 1137, 1140, 1142.