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If n is a positive integer and the product of all the integers from 1 to n, inclusive, is multiple of 990, what is the least possible value of n?
To solve this problem, we need to consider the prime factorization of 990. The prime factorization of 990 is 2 * 3^2 * 5 * 11.
For the product of all the integers from 1 to n to be a multiple of 990, it must have at least one factor of 2, two factors of 3, one factor of 5, and one factor of 11.
The least possible value of n for which this is true is n = 11. This is because 11! is the smallest factorial that has at least one factor of 2, two factors of 3, one factor of 5, and one factor of 11.
Therefore, the answer is 11.
