If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n ?
a. 10b. 11c. 12d. 13e. 14

Option B.

Step-by-step explanation:

n is a positive integer and product of all integers from 1 to n is a multiple of 990.

That means n! = 990 × k

Or n! = 2 × 5 × 3 × 3 × 11 × k

If n = 2, then 2! will not be a multiple of 990.

If n = 3, then 3! will not be a multiple of 990.

If n = 5, then 5! will not be a multiple of 990.

If n = 9, then 9! will not be a multiple of 990 because 11 will be missing.

Now if we take n = 11, then 11! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 = 39916800 which is divisible by 990.

[As we can see all multiples of 990 = 2, 5, 9, 11 are present in 11! ]

Therefore, option B. 11 will be the answer.

no solution

step-by-step explanation:

there is not enough information in the problem to solve.

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step-by-step explanation: