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 nowhere else to go.
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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

Problem

Let $S$ be a set of 6 integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$, then $b$ is not a multiple of $a$. What is the least possible value of an element in $S?$

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$

Solution

If we start with $1$, we can include nothing else, so that won't work.

If we start with $2$, we would have to include every odd number except $1$ to fill out the set, but then $3$ and $9$ would violate the rule, so that won't work.

Experimentation with $3$ shows it's likewise impossible. You can include $7$, $11$, and either $5$ or $10$ (which are always safe). But after adding either $4$ or $8$ we have no more places to go.

Finally, starting with $4$, we find that the sequence $4,5,6,7,9,11$ works, giving us $\boxed{\textbf{(C)} \text{ 4}}$. (Random_Guy)

Solution 2

We know that all the odd numbers (except 1) can be used.

$3, 5, 7, 9, 11$

Now we have 7 to choose from for the last number (out of $1, 2, 4, 6, 8, 10, 12$). We can eliminate 1, 2, 10, and 12, and we have $4, 6, 8$ 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 $9$, none of the numbers would work, but if we take out $3$, we get:

$4, 5, 6, 7, 9, 11$

So the least number is $4$, so the answer is $\boxed{\textbf{(C)} \text{ 4}}$.

-Baolan

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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