Here's the solution for question 2(a). The problem states that the LCM of $12$, $48$, and $m$ is $336$, and their GCF is $4$. We need to find the possible value(s) of $m$. Step 1: Write the prime factorization of the given numbers, the LCM, and the GCF. $$12 = 2^2 \times 3^1$$ $$48 = 2^4 \times 3^1$$ $$\text{LCM}(12, 48, m) = 336 = 2^4 \times 3^1 \times 7^1$$ $$\text{GCF}(12, 48, m) = 4 = 2^2$$ Let the prime factorization of $m$ be $2^x \times 3^y \times 7^z$. Step 2: Use the GCF to determine constraints on the exponents $x, y, z$. For the GCF, the exponent of each prime factor is the minimum of the exponents in $12$, $48$, and $m$. For prime factor $2$: $\min(2, 4, x) = 2$. This implies $x \ge 2$. For prime factor $3$: $\min(1, 1, y) = 0$ (since $3^0$ is in the GCF). This implies $y = 0$. For prime factor $7$: $\min(0, 0, z) = 0$ (since $7^0$ is in the GCF). This implies $z \ge 0$. Step 3: Use the LCM to determine constraints on the exponents $x, y, z$. For the LCM, the exponent of each prime factor is the maximum of the exponents in $12$, $48$, and $m$. For prime factor $2$: $\max(2, 4, x) = 4$. This implies $x \le 4$. For prime factor $3$: $\max(1, 1, y) = 1$. This implies $y \le 1$. For prime factor $7$: $\max(0, 0, z) = 1$. This implies $z = 1$.