Difference between revisions of "2018 AMC 10A Problems/Problem 17"
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| − | It will be safe to start with 5 or 6 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. | + | It will be safe to start with <math>5</math> or <math>6</math> 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 <math>4,</math> we can get <math>4,5,6,7,9,11.</math> So <math>4</math> works. | Trying <math>4,</math> we can get <math>4,5,6,7,9,11.</math> So <math>4</math> works. | ||
Revision as of 18:05, 27 August 2021
- The following problem is from both the 2018 AMC 12A #12 and 2018 AMC 10A #17, so both problems redirect to this page.
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
If we start with 
we can include nothing else, so that won't work. (Also note that 
is not an answer choice)
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 no more valid numbers.
Finally, starting with 
we find that the sequence 
works, giving us 
Solution 2
We know that all the odd numbers except namely 
can be used.
Now we have 
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
Video Solution
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. 
