You Didn't Tell Me Any Of These Numbers!

A set of 6 consecutive integers always contains:

• exactly 3 even numbers, hence their product has to be divisible by 8 (and also by 16, as at least one of these even numbers is divisible by 4 as well)

• exactly 2 numbers, which are divisible by 3, therefore their product has to be divisible by 8

• at least one (and at most 2) number, which is divisible by 5 and since we have 3 even numbers in the set as well, their product has to be divisible by 10

• we do not necessarily have a number in the set, which is divisible by 7, e. g. the set {1, 2, 3, 4, 5, 6} does not contain any and therefore neither is their product divisible by 7 (in general, we can take the set {7k + 1, 7k +2, 7k + 3, 7k + 4, 7k + 5, 7k + 6} as counterexample)

$\text {Hence, our answer is: } \boxed {7}$

Product of n consecutive numbers is always divisible by n!.So,here 6 consecutive numbers is divisible by 6!=720 which has factors like 8,9,10 but 7 is not there. So our answer is 7