Difference between revisions of "2018 AMC 10A Problems/Problem 17"

(Solution 2)
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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>

Revision as of 23:09, 4 February 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

Solution 3

We can get the multiples for the numbers in the original set with multiples in the same original set

$1:$ $\text{all}$ $\text{numbers}$ $\text{within}$ $\text{range}$

$2: 4,6,8,10,12$

$3: 6,9,12$

$4: 8,12$

$5: 10$

$6: 12$

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 $4$, we can get $4,5,6,7,9,11$. So $4$ works. Trying $3$ won't work, so the least is $4$. This means the answer is $\boxed{\textbf{(C) } 4}$

~OlutosinNGA

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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