In this aptitude odd-man-out question, select the odd number from the given alternatives.

Introduction / Context:
This is a basic aptitude odd-man-out question involving simple divisibility rules. You are given four two-digit numbers, and your task is to find the one that does not share a particular common property with the other three. Such questions check your number sense and your familiarity with divisibility by common factors like 3 and 5.

Given Data / Assumptions:

• Numbers given: 45, 56, 55 and 78.
• We assume standard divisibility rules taught in school mathematics.
• We are looking for a property that three numbers share and one number does not.

Concept / Approach:
A natural way to tackle odd-man-out questions with numbers is to test divisibility by small primes such as 2, 3, 5, 7, 11 and so on. We want to find a single property that holds for exactly three of these numbers and fails for one of them. Among common exam patterns, divisibility by 3 or 5 appears frequently, because these rules are easy to verify using the digits of the numbers.

Step-by-Step Solution:
Step 1: Check divisibility by 5. A number is divisible by 5 if it ends with 0 or 5. Here, 45 and 55 end with 5, so they are divisible by 5. The numbers 56 and 78 do not end with 0 or 5, so they are not divisible by 5. Step 2: Check divisibility by 3. A number is divisible by 3 if the sum of its digits is divisible by 3. For 45, the digit sum is 4 + 5 = 9, which is divisible by 3. For 55, the sum is 5 + 5 = 10, which is not divisible by 3. For 56, the sum is 5 + 6 = 11, which is not divisible by 3. For 78, the sum is 7 + 8 = 15, which is divisible by 3. Step 3: Summarise the results. 45 is divisible by both 3 and 5. 55 is divisible by 5 but not by 3. 78 is divisible by 3 but not by 5. 56 is divisible by neither 3 nor 5. Step 4: Identify the common property. The numbers 45, 55 and 78 are all divisible by either 3 or 5. The number 56 is the only one that is not divisible by 3 and not divisible by 5. Step 5: Conclude that 56 is the odd number because it fails this simple divisibility property that the other three satisfy.

Verification / Alternative check:
We can double check by considering other possible patterns such as prime factors, parity (odd or even) or relationships with squares. Although 24 and 78 are even and 45 and 55 are odd, this does not create a 3 versus 1 split. Similarly, all four numbers are composite. The only neat and unique pattern here is the one based on divisibility by 3 or 5, which clearly isolates 56.

Why Other Options Are Wrong:
45 is not odd-man-out because it is divisible by 3 and 5, so it fits the rule that the number is divisible by at least one of 3 or 5. 55 also fits because it is divisible by 5. 78 fits because it is divisible by 3. Only 56 does not satisfy divisibility by 3 or 5, so the other three are part of the same group and are not the answer.

Common Pitfalls:
A common mistake is to focus only on whether the numbers are odd or even and quickly choose any even number as the answer. However, in this set, 56 and 78 are both even, so parity alone cannot isolate a single odd-man-out. Another mistake is to look for overly complex patterns when a simple divisibility rule is sufficient. Always test small primes first, because exam questions usually rely on straightforward and elegant properties.

Final Answer:
The odd number in the list is 56, because it is the only number that is not divisible by 3 or 5, while the other three numbers are divisible by at least one of these two common factors.