Consider all two digit numbers which remain the same when their digits are interchanged (for example, 11, 22, 33, and so on). What is the average of all such two digit numbers?

Introduction / Context:
This question deals with a special set of two digit numbers whose digits are the same, meaning they are unchanged when the digits are interchanged. These are simple palindromic two digit numbers, and you are asked to find their average. The problem tests your ability to identify the full set correctly and then use a basic property of evenly spaced numbers.

Given Data / Assumptions:

• We consider two digit numbers where both digits are the same.
• Examples given include 11, 22, 33, and so on.
• We must find the average of all such numbers.
• All such numbers lie between 10 and 99.

Concept / Approach:
The numbers described are 11, 22, 33, 44, 55, 66, 77, 88 and 99. These form an arithmetic progression with a constant difference of 11 between consecutive terms. For any arithmetic progression, the average (mean) of all terms is equal to the average of the first and last terms. This allows a quick calculation without summing each term individually.

Step-by-Step Solution:
Step 1: List all two digit numbers with identical digits: 11, 22, 33, 44, 55, 66, 77, 88, 99. Step 2: Confirm that these form an arithmetic sequence with common difference 11. Step 3: The first term a = 11, the last term l = 99. Step 4: The average of numbers in an arithmetic progression is (a + l) / 2. Step 5: Compute (11 + 99) / 2 = 110 / 2 = 55. Step 6: Therefore, the average of all such two digit numbers is 55.

Verification / Alternative check:
We can verify by explicitly summing the numbers. Sum = 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99. Group them in pairs around the centre: (11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) + 55. Each pair sums to 110, and there are 4 such pairs plus the middle term 55. So total sum = 4 * 110 + 55 = 440 + 55 = 495. There are 9 numbers, so average = 495 / 9 = 55. This matches our earlier calculation.

Why Other Options Are Wrong:
Values such as 53, 54, 56 or 50 do not reflect the symmetric placement of the numbers around 55. Because the smallest term (11) and largest term (99) are equally spaced around 55, and the sequence is perfectly symmetric, the average must be exactly 55. Any other value would break this symmetry and would not equal (11 + 99) / 2.

Common Pitfalls:
Sometimes students mistakenly include numbers like 10 or 100 or forget to include all nine numbers. Others may incorrectly think they need to find the average of individual digits rather than the numbers themselves. Recognising the structure as a simple arithmetic progression and using the first and last term formula is the quickest and safest method.

Final Answer:
The average of all such two digit numbers is 55.