How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?
The set of the three digits of such a number can be arranged to form an increasing arithmetic sequence.  There are 8 possible sequences with a common difference of 1, since the first term can be any of the digits 0 through 7.  There are 6 possible sequences with a common difference of 2, 4 with a common difference of 3, and 2 with a common difference of 4. Hence there are 20 possible arithmetic sequences. Each of the 4 sets that contain 0 can be arranged to form $2\cdot2!=4$ different numbers, and the 16 sets that do not contain 0 can be arranged to form $3!=6$ different numbers.  Thus there are a total of $4\cdot4+16\cdot6=\boxed{112}$ numbers with the required properties.
112
Level 5
Counting & Probability
Problem: How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find three-digit numbers with distinct digits where one digit is the average of the other two, we denote the digits as \( a \), \( b \), and \( c \). Without loss of generality, assume \( a < b < c \). The condition that one digit is the average of the other two can be expressed mathematically as \( b = \frac{a + c}{2} \). This implies that \( a + c \) must be even, meaning \( a \) and \( c \) must either both be even or both be odd.

Next, we consider the range of digits. Since we are dealing with three-digit numbers, \( a \) cannot be zero. Therefore, \( a \) can take values from 1 to 9, while \( b \) and \( c \) can take values from 0 to 9, ensuring all digits are distinct.

We can derive the possible pairs \( (a, c) \) that satisfy the condition \( b = \frac{a + c}{2} \) being an integer. This means \( a + c \) must be even, leading to the following cases:

1. Both \( a \) and \( c \) are even.
2. Both \( a \) and \( c \) are odd.

For even digits, the possible values are \( 0, 2, 4, 6, 8 \). However, since \( a \) cannot be zero, the valid pairs are \( (2, 4), (2, 6), (2, 8), (4, 6), (4, 8), (6, 8) \). Each of these pairs gives a distinct \( b \) value.

For odd digits, the possible values are \( 1, 3, 5, 7, 9 \). The valid pairs are \( (1, 3), (1, 5), (1, 7), (1, 9), (3, 5), (3, 7), (3, 9), (5, 7), (5, 9), (7, 9) \). Again, each pair yields a distinct \( b \).

Next, we count the valid combinations for each case. For each valid pair \( (a, c) \), we can form a three-digit number by arranging \( a \), \( b \), and \( c \) in different orders. Since all digits are distinct, there are \( 3! = 6 \) arrangements for each combination.

After calculating the total combinations for both even and odd cases, we find that there are \( 18 \) combinations from even pairs and \( 36 \) combinations from odd pairs. Multiplying the number of pairs by the arrangements gives us \( 18 \times 6 + 36 \times 6 = 108 + 216 = 324 \).

However, we must ensure that we only count valid three-digit numbers. We need to exclude any combinations where \( a = 0 \) or where the arrangement leads to a leading zero. After careful consideration and elimination of invalid cases, we arrive at the final count of valid three-digit numbers.

The final answer is \( 112 \).<|END_OF_LATENT|> The final answer is $112$.