To solve the problem where we need to find the minimum possible value of a 3-digit number that, when divided by 7, 5, or 4, leaves a remainder of 2, we can approach it step by step.

Step 1: Understand the problem

We are asked to find a 3-digit number $n$ such that: $n \equiv 2 \pmod{7}, \quad n \equiv 2 \pmod{5}, \quad n \equiv 2 \pmod{4}$

This means $n - 2$ must be divisible by 7, 5, and 4.

Step 2: Find the Least Common Multiple (LCM)

The number $n - 2$ should be a multiple of the LCM of 7, 5, and 4. First, we find the LCM:

$\text{LCM}(7, 5, 4) = 7 \times 5 \times 4 = 140$

So, $n - 2 = 140k$ for some integer $k$, and thus: $n = 140k + 2$

Step 3: Find the minimum 3-digit $n$

We now want to find the smallest $k$ such that $n = 140k + 2$ is a 3-digit number.

To do that, $n \geq 100$, so: $140k + 2 \geq 100$ $140k \geq 98$ $k \geq \frac{98}{140} \approx 0.7$ Since $k$ must be an integer, the smallest $k$ is $k = 1$.

Step 4: Calculate $n$

For $k = 1$: $n = 140(1) + 2 = 142$

Therefore, the minimum possible value of $n$ is 142, which corresponds to option (C) 142.

Would you like more details or have any questions on this?

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Here are 5 related questions you might find helpful:

1. What would be the next possible 3-digit number after 142 that satisfies these conditions?
2. How would you solve this problem using the Chinese Remainder Theorem?
3. Can you generalize this method to any modulus, not just 7, 5, and 4?
4. What if the remainder was 3 instead of 2, how would the solution change?
5. How does the LCM affect the range of possible solutions?

Tip: Understanding the concept of modular arithmetic is key when dealing with divisibility and remainder problems.