1. **State the problem:** We are given two numbers expressed in prime factorization form: $$A = 2^2 \times 3^n \times 5$$ $$B = 2^m \times 3^2 \times 5^2$$ and their Highest Common Factor (HCF) is: $$\text{HCF}(A,B) = 2^2 \times 3^1 \times 5^1 = 60$$ We need to find the values of $m$ and $n$. 2. **Recall the formula for HCF using prime factors:** The HCF of two numbers is the product of the primes raised to the minimum power of each prime appearing in both numbers. $$\text{HCF}(A,B) = 2^{\min(2,m)} \times 3^{\min(n,2)} \times 5^{\min(1,2)}$$ 3. **Set up equations from the given HCF:** From the HCF given, we have: - For prime 2: $\min(2,m) = 2$ - For prime 3: $\min(n,2) = 1$ - For prime 5: $\min(1,2) = 1$ 4. **Solve for $m$ and $n$:** - Since $\min(2,m) = 2$, $m$ must be at least 2. So $m \geq 2$. - Since $\min(n,2) = 1$, $n$ must be 1 or greater but less than 2. So $n = 1$. 5. **Final values:** $$m \geq 2$$ $$n = 1$$ Since $m$ is an exponent in the factorization of $B$, the smallest integer satisfying $m \geq 2$ is $m=2$. **Answer:** $$m = 2, \quad n = 1$$