Concept:This is a digit puzzle based on place value and divisibility rules. We check each digit condition step by step to find the correct number.
Explanation:We are looking for a 5-digit number with all different digits that is divisible by 2 (last digit even) but not by 4 (last two digits not divisible by 4).
- Tens place must be the smallest prime number, which is 2.
- Hundreds place must be the second odd number. Odd numbers: 1, 3, 5… So second odd is 3.
- Thousands place must be the third prime number. Primes: 2, 3, 5, 7… So third prime is 5.
- Ten thousands place must be double of the tens place. Double of 2 is 4.
Thus, the digits from left to right: ten thousands = 4, thousands = 5, hundreds = 3, tens = 2. The units digit must be different and make the number divisible by 2 but not by 4. Check option C: 45326 has units digit 6 (even), last two digits 26 not divisible by 4, and all conditions match.
Option A: 25310 has tens digit 1, not 2. Option B: 26314 has tens digit 1, not 2. Option D: 46320 has ten thousands digit 4 but tens digit 2? Actually 46320 has tens digit 2, but ten thousands is 4 which is double of 2 – wait, check: 46320 – ten thousands place is 4 (double of 2) – but hundreds digit? Hundreds is 3? Actually 46320 digits: 4,6,3,2,0. Hundreds is 3 (correct), thousands is 6 (not 5), so violates thousands condition. So only C works.
Answer:C. 45326