Choose the correct answer.Given that the HCF of 2352 and 2a × 33 × 52 × 7b is 588, find the smallest possible values of a and b.a = 3 and b = 6a = 2 and b = 2a = 2 and b = 4a = 1 and b = 4

The HCF (Highest Common Factor) of two numbers is the product of the smallest power of each common prime factor present in the numbers.

The prime factorization of 2352 is 2^3 × 3^2 × 7^2 and the prime factorization of 2a × 3^3 × 5^2 × 7^b is 2^a × 3^3 × 5^2 × 7^b.

The HCF is given as 588, whose prime factorization is 2^2 × 3^0 × 5^0 × 7^2.

Comparing the powers of the prime factors in the HCF and the numbers, we get:

For 2: a = 2 (since 2^2 is in the HCF and 2^3 is in 2352, so we take the smaller power) For 3: there is no power of 3 in the HCF, so it doesn't affect the value of a. For 5: there is no power of 5 in the HCF, so it doesn't affect the value of a. For 7: b = 2 (since 7^2 is in the HCF and 7^2 is in 2352, so we take the smaller power)

So, the smallest possible values of a and b are a = 2 and b = 2. Therefore, the correct answer is "a = 2 and b = 2".