Even numbers are formed by taking at least two at a time from the numbers $0, 4, 8, 9$. Their H.C.F. is

### Concept & Logic The problem asks for the HCF of *all possible* even numbers formed by combining at least two digits from the set $\{0, 4, 8, 9\}$. The HCF is the largest number that divides every single valid number in that set. If we can find just a few valid numbers whose HCF is small, we can rule out larger options immediately. ### Step-by-Step Solution * **Given:** Digits to use: $0, 4, 8, 9$. Rule: Form even numbers using at least two digits. * **Calculation / Deduction:** 1. To be an even number, the last digit must be $0, 4$, or $8$. 2. Let's construct a few easy two-digit even numbers from this set to test the options. * Let's form $40$. * Let's form $48$. * Let's form $90$. * Let's form $94$. 3. Now, evaluate the HCF of just these selected examples: $40, 48, 90, 94$. 4. By definition, since they are all even, $2$ is a common factor. 5. Let's check if a higher number like $4$ is a common factor. Look at $90$. $90 \div 4 = 22.5$. Since $4$ does not divide $90$, $4$ cannot be the HCF of the entire set of numbers. 6. Let's check $10$. Look at $94$. $94$ does not end in $0$, so it is not divisible by $10$. 7. Therefore, the highest common factor that divides all the formed even numbers is $2$. ### Exam Strategy & Shortcut **Boundary Testing:** In questions involving "all possible numbers formed", purposely try to create numbers that break the given options. You want to eliminate the larger options ($4$ and $10$). To eliminate $10$, form an even number that doesn't end in $0$: $94$. To eliminate $4$, form an even number not divisible by $4$: $90$ or $98$. Once you have $90$ and $94$ in your valid set, their HCF is clearly $2$. This immediately leaves $2$ as the only possible answer. ### Common Pitfall Assuming that because $4$ and $8$ are in the digit pool, the resulting numbers will heavily favor multiples of $4$. Students often forget they can start a number with $9$ (creating numbers like $90, 94, 98$), which breaks the divisibility by $4$. ### Final Answer **Therefore, the correct answer is $2$.**