2018 AMC 10A Problems/Problem 17
- The following problem is from both the 2018 AMC 10A #17 and 2018 AMC 12A #12, so both problems redirect to this page.
Contents
Problem
Let be a set of 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 1
We start with because is not an answer choice. 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 no more valid numbers.
Finally, starting with we find that the sequence works, giving us
Solution 2
We know that all odd numbers except namely can be used.
Now we have possibilities to choose from for the last number (out of ). We can eliminate and and we have to choose from. However, is a multiple of Now we have to take out either or from the list. If we take out none of the numbers would work, but if we take out we get The least number is so the answer is
Solution 3
We can get the multiples for the numbers in the original set with multiples in the same original set It will be safe to start with or since they have the smallest number of multiples as listed above, but since the question asks for the least, it will be better to try others.
Trying we can get So works. Trying it won't work, so the least is This means the answer is
Solution 4
We partition into six nonempty subsets such that for every subset, each element is a multiple of all elements less than or equal to itself: Clearly, must contain exactly one element from each subset:
- For we can select either or
- For we can select either or Recall that since we cannot select
- For we can select either (provided that ) or Recall that since either or we can select neither nor
If then the possibilities of are or So, the least possible value of an element in is
Remark
There exist multiple such partitions of into six nonempty subsets, one of which is Regardless of which partition we use, we will conclude that to minimize the least element of the only possibilities for are or
~MRENTHUSIASM
Solution 5
We start with 2 as 1 is not an answer option. Our set would be . We realize we cannot add 12 to the set because 12 is a multiple of 3. Our set only has 5 elements, so starting with 2 won't work.
We try 3 next. Our set becomes . We run into the same issue as before so starting with 3 won't work.
We then try 4. Our set becomes . We see we have 6 elements with none being multiples of each other. Therefore our answer is
Video Solution
~IceMatrix
Video Solution
~savannahsolver
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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