Difference between revisions of "2018 AMC 10A Problems/Problem 17"
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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. | ||
− | Experimentation with <math>3</math> shows it's likewise impossible. You can include <math>7</math>, <math>11</math>, and either <math>5</math> or <math>10</math> (which are always safe). But after adding either <math>4</math> or <math>8</math> we have | + | Experimentation with <math>3</math> shows it's likewise impossible. You can include <math>7</math>, <math>11</math>, and either <math>5</math> or <math>10</math> (which are always safe). But after adding either <math>4</math> or <math>8</math> we have no more places to go. |
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>. |
Revision as of 22:13, 23 January 2019
Contents
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
If we start with , we can include nothing else, so that won't work.
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 . (Random_Guy)
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 .
-Baolan
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.