A number $x$ is divided by 7. When this number is divided by 8, 12 and 16. It leaves a remainder 3 in each case. The least value of $x$ is:

## Concept & Strategy To find a number that leaves the same remainder when divided by multiple divisors, we use the LCM (Least Common Multiple) of those divisors. The general form of the number is $N = \text{LCM}(a, b, c) \times k + R$. We then test integer values of $k$ to satisfy the secondary condition (divisibility by 7). ## Step-by-Step Solution * **Given:** Divisors are 8, 12, and 16. Remainder $R = 3$. The number must be a multiple of 7 (implied by "divided by 7" without a remainder, a common phrasing quirk for "divisible by 7" in exams). * **Step 1:** Find the LCM of 8, 12, and 16. * $8 = 2^3$ * $12 = 2^2 \times 3$ * $16 = 2^4$ * $\text{LCM}(8, 12, 16) = 2^4 \times 3 = 48$ * **Step 2:** Express the number in the general form. * Number $x = 48k + 3$ * **Step 3:** Find the least value of $k$ for which $x$ is divisible by 7. * For $k = 1$: $x = 48(1) + 3 = 51$ (Not divisible by 7) * For $k = 2$: $x = 48(2) + 3 = 99$ (Not divisible by 7) * For $k = 3$: $x = 48(3) + 3 = 147$. And $147 \div 7 = 21$. (Divisible by 7) ## Exam Strategy & Shortcut You can completely bypass the LCM calculation by using **Option Elimination**. The question states the number is divisible by 7. Test the given options directly: * (a) $148 \div 7 = 21.14$ (No) * (b) $149 \div 7 = 21.28$ (No) * (c) $150 \div 7 = 21.42$ (No) * (d) $147 \div 7 = 21$ (Yes) Only 147 is a perfect multiple of 7. ## Common Pitfall Students often find the LCM (48) and just add 3 to get 51, forgetting to check the divisibility condition for 7. Always verify that your final answer satisfies *all* constraints given in the problem. ## Final Answer Therefore, the correct answer is 147.