Difference between revisions of "2018 AMC 10A Problems/Problem 17"
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| + | {{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #12]] and [[2018 AMC 10A Problems|2018 AMC 10A #17]]}} | ||
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== Problem == | == Problem == | ||
Let <math>S</math> be a set of 6 integers taken from <math>\{1,2,\dots,12\}</math> with the property that if <math>a</math> and <math>b</math> are elements of <math>S</math> with <math>a<b</math>, then <math>b</math> is not a multiple of <math>a</math>. What is the least possible value of an element in <math>S?</math> | Let <math>S</math> be a set of 6 integers taken from <math>\{1,2,\dots,12\}</math> with the property that if <math>a</math> and <math>b</math> are elements of <math>S</math> with <math>a<b</math>, then <math>b</math> is not a multiple of <math>a</math>. What is the least possible value of an element in <math>S?</math> | ||
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<math>\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7</math> | <math>\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7</math> | ||
| − | == Solution == | + | == Solution 1 == |
| − | If we start with <math>1</math>, we can include nothing else, so that won't work. | + | If we start with <math>1</math>, we can include nothing else, so that won't work. (Also note that <math>1</math> is not an answer choice) |
If we start with <math>2</math>, we would have to include every odd number except <math>1</math> to fill out the set, but then <math>3</math> and <math>9</math> would violate the rule, so that won't work. | If we start with <math>2</math>, we would have to include every odd number except <math>1</math> to fill out the set, but then <math>3</math> and <math>9</math> would violate the rule, so that won't work. | ||
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Finally, starting with <math>4</math>, we find that the sequence <math>4,5,6,7,9,11</math> works, giving us <math>\boxed{\textbf{(C)} \text{ 4}}</math>. | Finally, starting with <math>4</math>, we find that the sequence <math>4,5,6,7,9,11</math> works, giving us <math>\boxed{\textbf{(C)} \text{ 4}}</math>. | ||
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==Solution 2== | ==Solution 2== | ||
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So the least number is <math>4</math>, so the answer is <math>\boxed{\textbf{(C)} \text{ 4}}</math>. | So the least number is <math>4</math>, so the answer is <math>\boxed{\textbf{(C)} \text{ 4}}</math>. | ||
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==Solution 3== | ==Solution 3== | ||
We can get the multiples for the numbers in the original set with multiples in the same original set | We can get the multiples for the numbers in the original set with multiples in the same original set | ||
| − | <math>1: all</math> <math>numbers</math> <math>within</math> <math>range</math> | + | <math>1:</math> <math>\text{all}</math> <math>\text{numbers}</math> <math>\text{within}</math> <math>\text{range}</math> |
<math>2: 4,6,8,10,12</math> | <math>2: 4,6,8,10,12</math> | ||
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Trying <math>3</math> won't work, so the least is <math>4</math>. This means the answer is <math>\boxed{\textbf{(C) } 4}</math> | Trying <math>3</math> won't work, so the least is <math>4</math>. This means the answer is <math>\boxed{\textbf{(C) } 4}</math> | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/M22S82Am2zM | ||
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==See Also == | ==See Also == | ||
Revision as of 17:50, 30 December 2020
- 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 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 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 places to go.
Finally, starting with , we find that the sequence 
works, giving us 
.
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 .
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 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.
Trying , we can get . So works.
Trying 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. 
