A number consists of two digits of which the ten's digit exceeds the unit digit by 7.The number itself is equal to 10 times the sum of the digits. Find the number

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.

A number consists of two digits of which the ten's digit exceeds the unit digit by 7. The number itself is equal to 10 times the sum of the digits. Find the number.

Solve a two-digit number problem where the ten's digit exceeds the unit digit by 7, and the number is equal to 10 times the sum of its digits. Learn key algebraic concepts and follow a step-by-step solution suitable for students in Grades 6-8.

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