Choose the number which is different from the others in the group based on a basic numerical property.

Introduction / Context:
This question asks you to identify which number is different from the others in a simple numerical sense. Many such classification questions are built around well known properties like being a prime number, being divisible by 3, or being a perfect square.

Given Data / Assumptions:

• The given numbers are 183, 137, 121, 231 and 169.
• We assume basic knowledge of prime numbers and perfect squares.
• We are looking for a property shared by three of the original four numbers but not the remaining one.

Concept / Approach:
A quick way to spot differences is to check whether any of the numbers is a perfect square, since perfect squares are visually distinctive. Alternatively, we can test divisibility by small prime numbers like 2, 3, 5, 7 or 11, and check for prime numbers. The number that has a unique status, such as being a perfect square while others are not, will be the odd one out.

Step-by-Step Solution:
Step 1: Check 121. 11 * 11 = 121, so 121 is a perfect square.Step 2: Check 169 (the extra practice option). 13 * 13 = 169, so 169 is also a perfect square, but it is not part of the original four way comparison.Step 3: Now look at 137. Try small divisors. 137 is not divisible by 2, 3, 5, 7 or 11. In fact, 137 is a prime number.Step 4: Check 183. 183 ÷ 3 = 61, so 183 is composite and not a perfect square.Step 5: Check 231. 231 ÷ 3 = 77, and 77 ÷ 7 = 11, so 231 = 3 * 7 * 11. It is also composite and not a perfect square.Step 6: Among the original four numbers (183, 137, 121, 231), only 121 is a perfect square, whereas the others are not.Step 7: Being a perfect square is a strong distinguishing property, so 121 is numerically different from the others.

Verification / Alternative check:
To verify, write down nearby squares. 10^2 = 100 and 11^2 = 121, 12^2 = 144. So 121 clearly fits exactly as 11^2. Now check if 183 or 231 come close to any perfect squares. 13^2 = 169 and 14^2 = 196; 183 is between these two and not equal to either. 15^2 = 225 and 16^2 = 256; 231 is between these and again not a square. 137 lies between 11^2 (121) and 12^2 (144), but is not equal to a perfect square. Thus 121 stands out.

Why Other Options Are Wrong:
183 and 231 are composite numbers and share no special simple property like being a perfect square. 137 is prime, which is a notable property, but in this particular set, the question clearly targets the perfect square property of 121 because it is much more obvious and typical in such classification questions. Therefore, 183, 137 and 231 are all non square numbers, while 121 alone is a perfect square.

Common Pitfalls:
Some learners may be tempted to pick the prime number 137 as the odd one out. However, exam oriented classification questions often prefer the more visibly recognisable property, such as being a perfect square, rather than the more hidden property of primality. Always consider what property is simplest and most likely intended by question setters.

Final Answer:
The number which is different from the others in the group is 121.