In the following number classification question, four numbers are given (145, 463, 581 and 651). Three of these are composite numbers, while one is a prime number. Select the odd number from the given alternatives.

Introduction / Context:
This question comes from the number theory and odd one out section. The numbers 145, 463, 581 and 651 must be checked for primality and factor structure. Three of them turn out to be composite numbers with non trivial factors, while one is a prime number with no divisors other than 1 and itself. Your task is to identify the prime number as the odd one out.

Given Data / Assumptions:

• Numbers to analyse: 145, 463, 581 and 651.
• Prime number: has exactly two positive divisors, 1 and itself.
• Composite number: has more than two positive divisors.
• We apply divisibility tests using small primes such as 2, 3, 5, 7, 11, 13, 17 and 19.
• Exactly one number in the list is prime.

Concept / Approach:
The concept is simple prime testing for three digit numbers. For each number, we try dividing by small primes up to roughly the square root of the number. If no divisor works, the number is prime. If we find a non trivial factor, the number is composite. By applying this method to all four options, we can see which one is prime and therefore the odd one out in a set dominated by composite numbers.

Step-by-Step Solution:
Step 1: Test 145.145 ends in 5, so it is divisible by 5. In fact, 145 = 5 * 29. Therefore, 145 is composite.Step 2: Test 581.Check divisibility by 7: 7 * 80 = 560, 7 * 83 = 581. Therefore, 581 = 7 * 83 and is composite.Step 3: Test 651.Check divisibility by 3: sum of digits is 6 + 5 + 1 = 12, which is divisible by 3. Hence 651 is divisible by 3. Indeed, 651 = 3 * 217, so it is composite.Step 4: Test 463.463 is not even, so not divisible by 2. Sum of digits is 4 + 6 + 3 = 13, not divisible by 3, so not divisible by 3. It does not end in 0 or 5, so not divisible by 5.Step 5: Continue testing 463 with primes up to its square root.Square root of 463 is a little more than 21. Try 7: 7 * 66 = 462, so 7 is not a divisor. Try 11: 11 * 42 = 462. Try 13: 13 * 35 = 455. Try 17: 17 * 27 = 459. Try 19: 19 * 24 = 456. None divide 463 exactly, so 463 has no small prime divisors and is prime.

Verification / Alternative check:
Once you have factored 145, 581 and 651, it is clear they are composite. At that point, 463 is the only candidate left to be prime. A quick double check confirms there is no easy factor: its last digit is 3, ruling out divisibility by 2 and 5, and the digit sum is 13, ruling out divisibility by 3. Testing common primes up to 21 secures that 463 has no divisors, completing the verification that it is prime.

Why Other Options Are Wrong:
145: Composite because 145 = 5 * 29.
581: Composite because 581 = 7 * 83.
651: Composite because 651 is divisible by 3 and can be factored further.

Common Pitfalls:
Students sometimes assume that any number ending in 1, 3, 7 or 9 is prime, which is not true. Many composite numbers, including 21 or 51, also end in these digits. In multiple choice tests, it is safer to run through quick divisibility tests and basic factorisation rather than relying on appearance. This habit ensures more reliable identification of primes and composites.

Final Answer:
The odd number is 463, because it is a prime number, whereas 145, 581 and 651 are composite numbers with non trivial factors.