If a number $774958A96B$ is divisible by 8 and 9, the respective values of $A$ and $B$ will be

### Concept & Logic For a number to be divisible by both 8 and 9, it must independently satisfy the divisibility rules for each. $$ \text{Rule for 8: The last three digits must be divisible by 8.} $$ $$ \text{Rule for 9: The sum of all digits must be divisible by 9.} $$ ### Step-by-Step Solution * **Given:** The number is $774958A96B$. * **Step 1: Apply the rule for 8.** The last three digits are $96B$. We need $\frac{96B}{8}$ to be an integer. * **Calculation:** 960 is exactly divisible by 8 ($960 \div 8 = 120$). The next multiple is 968. Therefore, the single digit $B$ can be either $0$ or $8$. * **Step 2: Apply the rule for 9 (Case 1: $B = 0$).** The number is $774958A960$. * Sum of digits $= 7+7+4+9+5+8+A+9+6+0 = 55 + A$. * For $55 + A$ to be divisible by 9, the next multiple of 9 is 63. * $55 + A = 63 \implies A = 8$. * This gives the pair $(A, B) = (8, 0)$. This matches option (c). * **Step 3: Check Case 2 ($B = 8$) just to be sure.** The number is $774958A968$. * Sum of digits $= 55 + 8 + A = 63 + A$. * For $63 + A$ to be divisible by 9, $A$ could be 0 or 9. * The pairs would be $(0, 8)$ or $(9, 8)$. Neither of these are in the options. ### Exam Strategy & Shortcut Use option elimination to save time. Since $B$ is the last digit, check the given options for $B$. Options provide $B=8, 8, 0$. Since both 0 and 8 work for divisibility by 8, you must check the sum for 9. By quickly substituting $B=0$ and $A=8$ (from option c), sum $= 55 + 8 = 63$, which is divisible by 9. You can verify the correct answer in seconds without solving for every possibility. ### Common Pitfall The most common mistake is miscalculating the large sum of digits. A great tip to avoid this is to "cast out nines". When adding $7+7+4+9+5+8+A+9+6+B$, simply cross out any 9s or combinations of digits that add to 9 (like $4+5$). This keeps your sum small and error-free! ### Final Answer Therefore, the correct answer is **8 and 0**.