Let the two-digit number be represented as $10a + b$, where $a$ is the ten's digit and $b$ is the unit digit.

We are given two conditions:

1. The ten's digit exceeds the unit digit by 7:
   $a = b + 7$

2. The number itself is equal to 10 times the sum of the digits:
   $10a + b = 10(a + b)$

Step 1: Simplify the second equation

We can expand and simplify the second condition: $10a + b = 10a + 10b$ Subtract $10a$ from both sides: $b = 10b$ Subtract $b$ from both sides: $0 = 9b$ Thus, $b = 0$.

Step 2: Substitute $b = 0$ into the first equation

From the first condition $a = b + 7$, substituting $b = 0$: $a = 0 + 7 = 7$

Step 3: Form the number

The number is $10a + b = 10(7) + 0 = 70$.

Thus, the number is $\boxed{70}$.

Would you like more details on any part of the solution?

Here are 5 related questions:

1. What if the ten's digit exceeds the unit digit by a different value?
2. How would the solution change if the sum of the digits was doubled?
3. Can you generalize a formula for such problems involving digit relations?
4. How many other two-digit numbers satisfy different similar conditions?
5. What happens if we reverse the digits of the number found?

Tip: Always check your solution by plugging values back into the original conditions to ensure accuracy.