Math Problem Statement

the sum of the digits of a 2 digit number is 8 The number obtained by interchanging the digits exceeds the original number by 18 Solve for the original number.

Solution

Let's solve the problem step by step:

Let the original two-digit number be represented as:

10a+b10a + b

where aa is the tens digit and bb is the ones digit.

Step 1: First condition - The sum of the digits is 8

The sum of the digits of the number is given as:

a + b = 8 \tag{1}

Step 2: Second condition - The number obtained by interchanging the digits exceeds the original number by 18

If the digits are interchanged, the new number becomes:

10b+a10b + a

According to the problem, this new number exceeds the original number by 18. So, we can write the equation:

10b+a=(10a+b)+1810b + a = (10a + b) + 18

Simplifying this equation:

10b+a=10a+b+1810b + a = 10a + b + 18

Now, subtract bb and aa from both sides:

9b9a=189b - 9a = 18

Dividing through by 9:

b - a = 2 \tag{2}

Step 3: Solve the system of equations

Now, we have the following system of equations:

  1. a+b=8a + b = 8
  2. ba=2b - a = 2

We can solve these equations by substitution or addition.

From equation (2), solve for bb:

b=a+2b = a + 2

Substitute this expression for bb into equation (1):

a+(a+2)=8a + (a + 2) = 8

Simplifying:

2a+2=82a + 2 = 8

Subtract 2 from both sides:

2a=62a = 6

Divide by 2:

a=3a = 3

Step 4: Find bb

Substitute a=3a = 3 into the equation b=a+2b = a + 2:

b=3+2=5b = 3 + 2 = 5

Step 5: Find the original number

The original number is:

10a+b=10(3)+5=3510a + b = 10(3) + 5 = 35

Thus, the original number is 35.

Verification:

  • The sum of the digits is 3+5=83 + 5 = 8, which satisfies the first condition.
  • Interchanging the digits gives 5353, and 5335=1853 - 35 = 18, which satisfies the second condition.

So, the original number is 35.


Would you like a more detailed breakdown of any part of the solution? Or do you have any other questions?

Here are some related questions for further exploration:

  1. How can we solve for a three-digit number where the sum of the digits is 15?
  2. What happens if the sum of the digits of a number is a negative value?
  3. How would you approach a problem where the difference between two numbers formed by digit interchange is given?
  4. How can you solve similar problems using algebraic expressions for three-digit or higher numbers?
  5. How do you handle problems where the sum or difference of digits is not an integer?

Tip: When solving problems involving digit manipulation, it's helpful to express the number as 10a+b10a + b and break the problem down into simpler equations based on the relationships between the digits.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Digit Manipulation

Formulas

10a + b
a + b = 8
10b + a = (10a + b) + 18

Theorems

System of Linear Equations

Suitable Grade Level

Grades 6-8