## 3. How many are there?

How many three-digit numbers are there for which the sum of two digits is twice the third?

## Result: $\quad 121$
## First Solution.

The desired property is possessed by all three-digit numbers with all identical digits, and there are 9 of them (precisely, these are $111,222,333, \ldots, 888,999$).

Now let's consider all combinations of three different digits regardless of their order, according to the condition of the problem. To ensure we list all possibilities, we will write them in ascending order, based on the difference between the digits.

| Difference between digits | Combinations | Number of combinations |
| :---: | :--- | :---: |
| 1 | $012,123,234,345,456,567,678,789$ | 8 |
| 2 | $024,135,246,357,468,579$ | 6 |
| 3 | $036,147,258,369$ | 4 |
| 4 | 048,159 | 2 |

The number of combinations with digits in ascending order is $8+6+4+2=20$.

In 16 combinations, there is no digit 0, and by choosing the order for each combination, we get 6 numbers (e.g., $123,132,213,231,312$ and 321). For the remaining 4 combinations where the digit 0 appears, there are only 4 possibilities, as the hundreds digit cannot be 0 (e.g., 102, 120, 201, and 210, but not 012 and 021).

The total number of such numbers with different digits is $16 \cdot 6 + 4 \cdot 4 = 96 + 16 = 112$.

The total number of desired numbers is $9 + 112 = 121$.