The difference of the squares of the digits of 
a 2-digit positive number is 27. If the digits 
are reversed in order and the resulting 
number is subtracted from the original 
number, the difference is also 27. What is 
the original number? 

Let's solve this step-by-step by breaking down the given conditions:

1. **Define the two-digit number:**

   Let the original number be \(10a + b\), where \(a\) is the tens digit and \(b\) is the units digit.

2. **Condition on the difference of the squares of the digits:**

   The problem states that the difference between the squares of the digits is 27:
   \[
   a^2 - b^2 = 27.
   \]

3. **Condition on reversing the digits:**

   If we reverse the digits, the new number becomes \(10b + a\). The difference between the original number and this reversed number is also 27:
   \[
   (10a + b) - (10b + a) = 27.
   \]

   Simplifying this equation:
   \[
   10a + b - 10b - a = 27,
   \]
   \[
   9a - 9b = 27.
   \]

   Dividing the entire equation by 9:
   \[
   a - b = 3.
   \]

4. **Solve the system of equations:**

   Now we have two equations:
   \[
   a^2 - b^2 = 27, \quad (1)
   \]
   \[
   a - b = 3. \quad (2)
   \]

   Notice that \(a^2 - b^2\) can be factored using the difference of squares formula:
   \[
   a^2 - b^2 = (a - b)(a + b).
   \]
   Substituting from equation (2), \(a - b = 3\):
   \[
   3(a + b) = 27.
   \]
   Dividing by 3:
   \[
   a + b = 9. \quad (3)
   \]

5. **Find \(a\) and \(b\):**

   Now we solve the system of linear equations given by (2) and (3):
   - \(a - b = 3\)
   - \(a + b = 9\)

   Adding these two equations:
   \[
   (a - b) + (a + b) = 3 + 9,
   \]
   \[
   2a = 12 \implies a = 6.
   \]

   Substituting \(a = 6\) into equation (2):
   \[
   6 - b = 3,
   \]
   \[
   b = 3.
   \]

6. **Conclusion:**

   The original number is \(10a + b = 10 \times 6 + 3 = 63\).

### Verification

- **Difference of squares:** \(a^2 - b^2 = 6^2 - 3^2 = 36 - 9 = 27\).  
- **Difference when digits are reversed:** Original number is 63, reversed is 36, and \(63 - 36 = 27\).

Both conditions are satisfied, so the original number is **63**.

---

Would you like further details, or do you have any questions?

Here are 5 related questions to further your understanding:

1. What would be the solution if the difference between the squares of the digits was 15 instead?
2. How would the solution change if the reversed number was subtracted from the original number to get a different result?
3. Can you think of a three-digit number where a similar problem could be applied?
4. What mathematical concept is primarily used to solve this type of problem?
5. How would you set up a system of equations for a similar problem with different conditions?

**Tip:** When solving problems involving digits, setting up equations based on place values and digit properties is a powerful technique.

Explore how to solve a problem involving a two-digit number where the difference of the squares of its digits is 27. Follow a detailed step-by-step solution involving equations and number theory concepts. Suitable for Grades 7-9.

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