A three digit number has 3 in the tens place. When the digits in the hundreds and units places are interchanged symmetrically about the tens position, the new number becomes 396 more than the original number. The sum of the digits in the units and hundreds places is 14. What is the original three digit number?

Introduction / Context:
This problem involves a three digit number and uses place value concepts along with basic algebra. The number is described in terms of its hundreds, tens and units digits, and a condition is given after interchanging two of the digits. Such questions test a learner understanding of how to represent numbers algebraically based on their digits and how to form and solve equations from word conditions.

Given Data / Assumptions:
- Let the original three digit number have hundreds digit a, tens digit 3 and units digit b. - The tens place is fixed as 3. - When the hundreds and units digits are interchanged, the new number is 396 more than the original number. - The sum of the hundreds and units digits is given as a + b = 14.

Concept / Approach:
- Use place value to express the original and new numbers in terms of a and b. - Write the algebraic equation that represents the condition that the new number is 396 more than the original number. - Combine this equation with the sum condition a + b = 14 to solve for the digits a and b. - Form the original number 100a + 30 + b and match with the options.

Step-by-Step Solution:
Step 1: Represent the original number as 100a + 30 + b. Step 2: After swapping the hundreds and units digits, the new number becomes 100b + 30 + a. Step 3: The new number is 396 more than the original, so write the equation: 100b + 30 + a = 100a + 30 + b + 396. Step 4: Simplify this equation. Subtract 30 from both sides and rearrange terms to get 100b + a = 100a + b + 396. Step 5: Bring like terms together: 100b - b = 100a - a + 396, which gives 99b = 99a + 396. Step 6: Divide every term by 99 to obtain b = a + 4. Step 7: Use the second condition a + b = 14. Substitute b = a + 4 to get a + (a + 4) = 14. Step 8: Simplify: 2a + 4 = 14 so 2a = 10 and a = 5. Then b = 5 + 4 = 9. Step 9: The original number is 100a + 30 + b = 100 * 5 + 30 + 9 = 539.

Verification / Alternative check:
Check the conditions with 539. The new number after swapping hundreds and units digits is 935. The difference 935 - 539 = 396, which matches the problem statement. The sum of the hundreds and units digits is 5 + 9 = 14, also matching the given condition. So the solution is fully consistent.

Why Other Options Are Wrong:
Option A (513): Here a + b = 5 + 3 = 8, not 14, so it fails the sum condition. Option C (439): The tens digit is 3, but a + b = 4 + 9 = 13 and the difference after swapping would not be 396. Option D (613): The tens digit is 1, which already violates the fact that the tens digit must be 3. Option E (731): Again, the tens digit is 3, but a + b = 7 + 1 = 8 and the required difference condition does not hold.

Common Pitfalls:
- Confusing the tens place with the hundreds or units place. - Forgetting that the tens digit is fixed and only the hundreds and units digits are swapped. - Making algebraic mistakes when simplifying the equation or solving the system for a and b.

Final Answer:
The original three digit number that satisfies all the conditions is 539.