In this number-classification question, select the odd four-digit number from the given alternatives.

Introduction / Context:
This is a straightforward numerical odd-man-out question. You are given four four-digit numbers, and exactly one of them is structurally different from the rest. Questions like this often use a simple visible property such as divisibility by 10, pattern of digits or multiples of a certain base. Your goal is to quickly test these obvious properties and find the one number that does not fit.

Given Data / Assumptions:

• Numbers given: 6210, 7020, 1431 and 1280.
• We use standard base 10 representation.
• We are looking for a common structural property that three numbers share and one does not share.

Concept / Approach:
One of the easiest structural properties to check with four-digit numbers is whether they end in zero. Numbers that end in zero are multiples of 10. Often, exam setters give three multiples of 10 and one number that is not a multiple of 10 to create a quick odd-man-out question. Therefore we start by checking the unit digit of each number, which requires minimal computation and gives an immediate pattern if it exists.

Step-by-Step Solution:
Step 1: Observe the unit digits. For 6210, the last digit is 0. For 7020, the last digit is again 0. For 1431, the last digit is 1. For 1280, the last digit is 0. Step 2: Identify which numbers are multiples of 10. A number is a multiple of 10 if it ends in 0. Therefore, 6210, 7020 and 1280 are all multiples of 10. The number 1431 ends in 1, so it is not a multiple of 10. Step 3: Check whether any other simple pattern could change our conclusion. For instance, 6210, 7020 and 1280 are even numbers and also divisible by 10. The number 1431 is odd and not divisible by 10. Thus 1431 clearly behaves differently. Step 4: Since three numbers are divisible by 10 and one is not, the one that is not divisible by 10 must be the odd-man-out under a clean and simple rule. Step 5: Conclude that 1431 is the odd number.

Verification / Alternative check:
To confirm, we can also examine other properties, such as the sum of digits. However, such checks are not necessary when we already have a strong and simple pattern that matches three numbers exactly. None of the alternative observations like being even or odd, or specific prime factorisations, provide as tidy a 3 versus 1 split as the divisibility by 10 property does. This strengthens our confidence that the intended classification is based on the last digit of the numbers, making 1431 the correct choice.

Why Other Options Are Wrong:
6210 is not odd because it ends with 0 and is a multiple of 10, just like 7020 and 1280. 7020 also ends with 0 and fits into the same group. 1280 ends with 0 and is a multiple of 10 as well. Therefore, these three numbers form a similar class. Only 1431 breaks this pattern because it ends with 1 and is not a multiple of 10, so the other three options are not valid odd-men-out choices.

Common Pitfalls:
Learners sometimes overcomplicate such questions by searching for hidden factor patterns or digit sum relationships. While such patterns can appear in some questions, in many classification items the examiner expects you to notice very simple features like ending digits, parity or basic divisibility. Ignoring these obvious checks and jumping straight into heavy calculations can waste time and still miss the correct pattern. The safe strategy is to first check simple visible properties such as last digit, number of digits and divisibility by 10 or 5.

Final Answer:
The odd number is 1431, because it is the only number that is not a multiple of 10, whereas 6210, 7020 and 1280 all end in zero and are divisible by 10.