2018 AMC 10A Problems/Problem 17

Problem

Let be a set of 6 integers taken from with the property that if and are elements of with , then is not a multiple of . What is the least possible value of an element in

Solution

If we start with , we can include nothing else, so that won't work.

If we start with , we would have to include every odd number except to fill out the set, but then and would violate the rule, so that won't work.

Experimentation with shows it's likewise impossible. You can include , , and either or (which are always safe). But after adding either or we have nowhere else to go.

Finally, starting with , we find that the sequence works, giving us . (Random_Guy)

Solution 2

We know that all the odd numbers (except 1) can be used.

Now we have 7 to choose from for the last number (out of ). We can eliminate 1, 2, 10, and 12, and we have to choose from. But wait, 9 is a multiple of 3! Now we have to take out either 3 or 9 from the list. If we take out , none of the numbers would work, but if we take out , we get:

So the least number is , so the answer is .

-Baolan

See Also

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