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Let's solve the problem.

Let the three-digit number be represented as \( \overline{abc} \), where:
- \( a \) is the hundreds digit,
- \( b \) is the tens digit,
- \( c \) is the units digit.

The problem gives us two conditions:
1. The units digit \( c \) is the sum of the hundreds and tens digits: 
   \[
   c = a + b
   \]
2. The units digit \( c \) exceeds the tens digit \( b \) by as much as the tens digit exceeds the hundreds digit:
   \[
   c - b = b - a
   \]
   This simplifies to:
   \[
   c = 2b - a
   \]

Now, we have the system of equations:
1. \( c = a + b \)
2. \( c = 2b - a \)

### Step 1: Set the two expressions for \( c \) equal to each other:
\[
a + b = 2b - a
\]

### Step 2: Solve for \( a \):
\[
a + b = 2b - a
\]
Add \( a \) to both sides:
\[
2a + b = 2b
\]
Subtract \( b \) from both sides:
\[
2a = b
\]
Thus, the tens digit \( b = 2a \).

### Step 3: Substitute \( b = 2a \) into the first equation for \( c \):
\[
c = a + b = a + 2a = 3a
\]

### Step 4: Now, since \( a \), \( b \), and \( c \) are digits, they must be integers between 0 and 9. Let's check possible values of \( a \):
- \( a = 1 \): \( b = 2a = 2 \), \( c = 3a = 3 \) → The number is \( 123 \).
- \( a = 2 \): \( b = 2a = 4 \), \( c = 3a = 6 \) → The number is \( 246 \).
- \( a = 3 \): \( b = 2a = 6 \), \( c = 3a = 9 \) → The number is \( 369 \).

For \( a = 4 \) or higher, \( b = 2a \) would be 8 or higher, and \( c = 3a \) would be greater than 9, which is not allowed since digits must be between 0 and 9.

### Conclusion:
There are 3 such three-digit numbers: \( 123 \), \( 246 \), and \( 369 \).

Thus, the answer is \( \boxed{3} \).

---

Would you like more details on any step? Here are 5 related questions to further explore this topic:

1. How would the solution change if the problem specified a different relationship between the digits?
2. Can this method be applied to other multi-digit numbers? If so, how?
3. What if the sum of the digits had to follow a different rule, such as being divisible by a certain number?
4. How could the concept of modular arithmetic help in solving this type of digit problem?
5. Can you find other problems where the difference between digits plays a crucial role in the solution?

**Tip:** When working with digit problems, always pay attention to the constraints on the values of the digits (i.e., they must be between 0 and 9).

In a three-digit number, the units digit is the sum of the other two digits and it exceeds the tens digit by as much as the latter exceeds the hundreds digit. How many such three-digit numbers exist?

Discover how to solve a three-digit number problem where the units digit is the sum of the other two digits, and how it relates to the tens and hundreds digits. Follow a step-by-step solution involving algebraic concepts and equations. Suitable for students in Grades 8-10.

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