Difference between revisions of "2018 AMC 10A Problems/Problem 17"
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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>. | ||
(Random_Guy) | (Random_Guy) | ||
+ | |||
+ | ==See Also == | ||
+ | {{AMC10 box|year=2018|ab=A|num-b=16|num-a=18}} | ||
+ | {{AMC12 box|year=2018|ab=A|num-b=11|num-a=13}} | ||
+ | {{MAA Notice}} |
Revision as of 17:33, 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.