In this number-classification odd-man-out question, select the number that does not follow the same divisibility pattern as the others.

Introduction / Context:
This is a numerical odd-man-out question where all the options are two or three digit integers. The task is to detect a property common to three of them and absent in one. Many such questions use divisibility by a fixed prime as the hidden theme. Here, checking divisibility by 17 reveals a neat and unique pattern that allows us to identify the odd number quickly.

Given Data / Assumptions:

• Numbers given: 86, 68, 136 and 102.
• We use standard integer arithmetic and divisibility.
• We aim to find one number that does not share a key divisor with the other three.

Concept / Approach:
A good way to approach such questions is to check for common divisors. Because the numbers are not extremely large, we can try divisibility by primes like 2, 3, 5, 7, 11, 13 and 17. Once we see that several numbers share a specific prime factor, we can verify whether one number is missing that factor. If so, that number is the odd-man-out. In this problem, all four numbers are even, so divisibility by 2 does not help, but checking divisibility by 17 is productive.

Step-by-Step Solution:
Step 1: Factor 68. Divide 68 by 17. We get 68 / 17 = 4, so 68 is a multiple of 17. Step 2: Factor 136. Divide 136 by 17. We get 136 / 17 = 8, so 136 is also a multiple of 17. Step 3: Factor 102. Divide 102 by 17. We get 102 / 17 = 6, so 102 is again a multiple of 17. Step 4: Check 86 for divisibility by 17. Divide 86 by 17. We get 86 / 17 = 5.0588..., which is not an integer. Therefore, 86 is not divisible by 17. Step 5: Summarise. We have discovered that 68, 136 and 102 are all divisible by 17, while 86 is not. Step 6: Conclude that since three numbers share the prime factor 17 and one does not, the number that is not divisible by 17 is the odd-man-out.

Verification / Alternative check:
We can look briefly at other properties such as multiples of 4, 8 or 3. For example, 68, 136 and 102 are all even, and so is 86, so parity does not help. Some of them are divisible by 4, but not all of them, and that does not create a clean 3 versus 1 split. The only elegant classification that clearly groups three numbers together is based on the factor 17. Therefore, any alternative scheme is either more contrived or fails to satisfy the odd-man-out condition cleanly.

Why Other Options Are Wrong:
68 is not the odd number because it is divisible by 17 and stands in the same class as 136 and 102. 136 is often noticed as a larger number, but size alone is not the basis of classification here; it is divisible by 17 just like the others. 102 also shares the factor 17. These three numbers therefore belong to one category. Only 86 fails the divisibility by 17 test and is consequently different from the rest.

Common Pitfalls:
A frequent error is to choose a number simply because it looks different, for example by being the smallest or largest, without checking arithmetic properties. Another mistake is to stop after checking divisibility by 2 or 3 and assuming there is no pattern. In exam situations, once you see that all numbers are even, it is a good idea to test slightly larger primes such as 7, 11, 13 or 17. This often reveals the intended classification, as it does very cleanly in this question.

Final Answer:
The odd number is 86, because 68, 136 and 102 are all divisible by 17, whereas 86 is not a multiple of 17.