If both 112 and 33 are factors of the number a x 43 x 62 x 1311, then what is the smallest possible value of 'a'?

To find the smallest possible value of 'a', we need to ensure that 'a' includes all prime factors of 112 and 33 that are not already included in 43, 62, and 1311.

Step 1: Prime factorize 112, 33, 43, 62, and 1311.

112 = 2^4 * 7 33 = 3 * 11 43 = 43 (it's a prime number) 62 = 2 * 31 1311 = 1311 (it's a prime number)

Step 2: Identify the prime factors of 'a' that are not already included in 43, 62, and 1311.

From the prime factorization, we can see that the prime factors of 112 and 33 are 2, 7, 3, and 11. The prime factors of 43, 62, and 1311 are 43, 2, 31, and 1311.

The prime factors 2 and 31 are already included in 62, so we don't need to include them in 'a'. However, the prime factors 7, 3, and 11 are not included in 43, 62, or 1311, so they must be included in 'a'.

Step 3: Calculate the smallest possible value of 'a'.

The smallest possible value of 'a' is the product of its prime factors, which are 7, 3, and 11.

So, a = 7 * 3 * 11 = 231.

Therefore, the smallest possible value of 'a' is 231.