Difference between revisions of "2018 AMC 10A Problems/Problem 17"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
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| + | ==Solution 5== | ||
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| + | We start with 2 as 1 is not an answer option. Our set would be <math>\{2,3,5,7,11\}</math>. 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. | ||
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| + | We try 3 next. Our set becomes <math>\{3,4,5,7,11\}</math>. We run into the same issue as before so starting with 3 won't work. | ||
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| + | We then try 4. Our set becomes <math>\{4,5,6,7,9,11\}</math>. We see we have 6 elements with none being multiples of each other. Therefore our answer is <math>\boxed{\textbf{(C)}\ 4}.</math> | ||
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| + | ~[https://artofproblemsolving.com/wiki/index.php/User:South South] | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 19:32, 11 June 2023
- 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 ![\[4, 5, 6, 7, 9, 11.\]](http://latex.artofproblemsolving.com/d/6/a/d6a9c1d25a9c8c6c44c0ea9c1db554e56e2211e1.png)
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: ![\[\{1,2,4,8\}, \ \{3,6,12\}, \ \{5,10\}, \ \{7\}, \ \{9\}, \ \{11\}.\]](http://latex.artofproblemsolving.com/a/1/b/a1b51d74e9b9f113aff4007008b3ba7e706b2741.png)
Clearly, must contain exactly one element from each subset:
- For
we can select either 
or 
we can select either 
- For
we can select either 
or 
Recall that since 
we cannot select 
we can select either 
Recall that since 
we cannot select - For
we can select either 
(provided that 
) or 
Recall that since either 
or 
we can select neither 
nor 
we can select either 
) or 
Recall that since either 
or 
we can select neither 
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 ![\[\{1,2,6,12\}, \ \{3,9\}, \ \{4,8\}, \ \{5,10\}, \ \{7\}, \ \{11\}.\]](http://latex.artofproblemsolving.com/a/a/0/aa06facdffeaa11b463d5349f7df4aa937417769.png)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
