Difference between revisions of "2018 AMC 10A Problems/Problem 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 values 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 values 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> | ||
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== Solution == | == Solution == | ||
Intuitively, one would see this list and start with prime numbers. However, there are only 5 prime numbers less than <math>12</math>, making this impossible. It is also clear that another number can't be added in, so <math>2</math> can't be the smallest. Next, we start the sequence with <math>3</math>, and a bit of trial and error shows it's impossible. Lastly, 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>. | Intuitively, one would see this list and start with prime numbers. However, there are only 5 prime numbers less than <math>12</math>, making this impossible. It is also clear that another number can't be added in, so <math>2</math> can't be the smallest. Next, we start the sequence with <math>3</math>, and a bit of trial and error shows it's impossible. Lastly, 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>. | ||
Revision as of 20:47, 8 February 2018
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 values of an element in 
Solution
Intuitively, one would see this list and start with prime numbers. However, there are only 5 prime numbers less than 
, making this impossible. It is also clear that another number can't be added in, so 
can't be the smallest. Next, we start the sequence with 
, and a bit of trial and error shows it's impossible. Lastly, starting with 
, we find that the sequence 
works, giving us 
.
(Random_Guy)
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. 
