Final answer:
To find the smallest integer k such that k! is divisible by 990, factorize 990 and ensure that k! includes the required prime factors with their multiplicities. The smallest value that satisfies this condition is k=12.
Step-by-step explanation:
The question asks to find the smallest positive integer k such that k! (k factorial) is divisible by 990. To solve this, we need to factorize 990 and understand the concept of factorial.
Firstly, we factor 990 into its prime factors, which gives us 990 = 2 × 32 × 5 × 11. To have a factorial divisible by 990, k! must include at least these prime factors in the necessary quantities.
By analyzing the prime factors required, we can deduce that k must be at least as large as the highest prime number in the factorization of 990, which is 11. However, we must also ensure that there are at least two factors of 3. Checking through factorials, we find that k = 12 meets these conditions because 12! includes each of the necessary prime factors and their multiplicities to be divisible by 990.
Therefore, the smallest positive integer k such that k! is divisible by 990 is 12.