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'?Options121363333267

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

Step 1: Prime factorize 112 and 33. 112 = 2^4 * 7 33 = 3 * 11

Step 2: Prime factorize 43, 62, and 1311. 43 = 43 (prime number) 62 = 2 * 31 1311 = 13 * 101

Step 3: Identify the missing prime factors. From the prime factorization, we can see that 'a' must be divisible by 2, 3, 7, and 11 to include all the prime factors of 112 and 33. However, 2 is already included in 62, so 'a' must be divisible by 3, 7, and 11.

Step 4: Find the smallest number that is divisible by 3, 7, and 11. The smallest number that is divisible by 3, 7, and 11 is their product, which is 231.

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