What is the greatest number of $3$ digits which when divided by $6$, $9$ and $12$ leaves a remainder of $3$ in each case?

### Concept & Logic To find the greatest $n$-digit number that leaves a remainder $R$ when divided by a set of numbers, you must follow a three-step process: 1. Find the LCM of the divisors. 2. Find the greatest $n$-digit multiple of that LCM. 3. Add the required remainder $R$ to that multiple. The general formula is: $$\text{Required Number} = (\text{Greatest } n\text{-digit Multiple of LCM}) + R$$ ### Step-by-Step Solution * **Given:** Divisors are $6$, $9$, and $12$. Remainder $R = 3$. We need the greatest $3$-digit number. * **Find the LCM:** $6 = 2 \times 3$ $9 = 3^2$ $12 = 2^2 \times 3$ $$LCM = 2^2 \times 3^2 = 4 \times 9 = 36$$ * **Find the greatest 3-digit multiple of the LCM:** The greatest general 3-digit number is $999$. Let's divide $999$ by our LCM ($36$) to find the closest exact multiple. $999 \div 36 = 27$ with a remainder of $27$. To make $999$ perfectly divisible by $36$, subtract this remainder: $$\text{Greatest 3-digit multiple} = 999 - 27 = 972$$ ($972$ is exactly $36 \times 27$). * **Add the required remainder:** The question specifies a remainder of $3$ in each case. $$\text{Required Number} = 972 + 3 = 975$$ ### Exam Strategy & Shortcut **Divisibility by 9 Reverse-Engineering:** The final number leaves a remainder of $3$ when divided by $9$. Therefore, if we subtract $3$ from the correct option, it MUST be perfectly divisible by $9$. (Sum of digits must equal a multiple of $9$). Let's test: (a) $903 - 3 = 900 \rightarrow$ Sum = $9$ (Possible) (b) $939 - 3 = 936 \rightarrow$ Sum = $18$ (Possible) (c) $975 - 3 = 972 \rightarrow$ Sum = $18$ (Possible) (d) $996 - 3 = 993 \rightarrow$ Sum = $21$ (Eliminated) Now use divisibility by $12$ (which means divisible by $4$). The number $(Option - 3)$ must be divisible by $4$ (last two digits divisible by $4$): (a) $900$: Last two digits $00$ (Divisible by $4$). (b) $936$: Last two digits $36$ (Divisible by $4$). (c) $972$: Last two digits $72$ (Divisible by $4$). Since the question asks for the **greatest** 3-digit number, we simply pick the largest valid option from our list, which is $975$. ### Common Pitfall The most frequent error is calculating the LCM ($36$), adding the remainder to get $39$, and then trying to find multiples of $39$ up to $999$. This is mathematically incorrect. You must find the multiple of the LCM first ($972$) and *then* add the remainder ($3$) at the very end. ### Final Answer **Therefore, the correct answer is 975.**