The greatest four digit number which is exactly divisible by each one of the numbers 12, 18, 21 and 28

## Concept & Strategy To find the greatest number of a certain length divisible by a set of numbers, we first find the Least Common Multiple (LCM) of those numbers. Any number divisible by the given set must be a multiple of their LCM. We then divide the maximum possible number of that length (e.g., 9999 for 4 digits) by the LCM and subtract the remainder to get our answer. ## Step-by-Step Solution * **Given:** Divisors are 12, 18, 21, and 28. The target is the greatest 4-digit number (which is $\le 9999$). * **Step 1:** Find the LCM of 12, 18, 21, and 28 using prime factorization. * $12 = 2^2 \times 3$ * $18 = 2 \times 3^2$ * $21 = 3 \times 7$ * $28 = 2^2 \times 7$ * $\text{LCM} = 2^2 \times 3^2 \times 7 = 4 \times 9 \times 7 = 252$ * **Step 2:** Divide the largest 4-digit number (9999) by the LCM (252). * $9999 \div 252 = 39$ with a remainder. * $252 \times 39 = 9828$ * $\text{Remainder} = 9999 - 9828 = 171$ * **Step 3:** Subtract the remainder from 9999 to find the greatest multiple of 252. $$9999 - 171 = 9828$$ ## Exam Strategy & Shortcut Use **Divisibility Rules** directly on the options to save time! The number must be divisible by 18 (which means it's divisible by 9) and 28 (which means it's divisible by 7 and 4). * Rule of 9 (sum of digits must be a multiple of 9): * (a) $9+2+8+8 = 27$ (Yes) * (b) $9+8+8+2 = 27$ (Yes) * (c) $9+8+2+8 = 27$ (Yes) * (d) $9+9+2+8 = 28$ (No) * Rule of 7 (divide by 7): * (b) $9882 \div 7$ leaves remainder 5 (No) * (a) $9288 \div 7$ leaves remainder 6 (No) * (c) $9828 \div 7 = 1404$ (Yes!) By quickly checking divisibility, you avoid calculating the LCM entirely. ## Common Pitfall Students often calculate the LCM (252) correctly but then try to manually multiply it up to 9000, wasting valuable time. Always use the division-and-remainder method ($9999 \div 252$) to jump straight to the correct multiplier. ## Final Answer Therefore, the correct answer is 9828.