Difference between revisions of "2018 AMC 10A Problems/Problem 17"
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+ | {{duplicate|[[2018 AMC 10A Problems/Problem 17|2018 AMC 10A #17]] and [[2018 AMC 12A Problems/Problem 12|2018 AMC 12A #12]]}} | ||
+ | |||
== 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 value of an element in <math>S | + | Let <math>S</math> be a set of <math>6</math> 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 value of an element in <math>S</math>? |
<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> | ||
− | == Solution == | + | == Solution 1 == |
− | + | We start with <math>2</math> because <math>1</math> is not an answer choice. 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>7,11,</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 valid numbers. | ||
+ | |||
+ | 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)}\ 4}.</math> | ||
− | + | ==Solution 2== | |
+ | We know that all odd numbers except <math>1,</math> namely <math>3, 5, 7, 9, 11,</math> can be used. | ||
− | + | Now we have <math>7</math> possibilities to choose from for the last number (out of <math>1, 2, 4, 6, 8, 10, 12</math>). We can eliminate <math>1, 2, 10,</math> and <math>12,</math> and we have <math>4, 6, 8</math> to choose from. However, <math>9</math> is a multiple of <math>3.</math> Now we have to take out either <math>3</math> or <math>9</math> from the list. If we take out <math>9,</math> none of the numbers would work, but if we take out <math>3,</math> we get <cmath>4, 5, 6, 7, 9, 11.</cmath> | |
+ | The least number is <math>4,</math> so the answer is <math>\boxed{\textbf{(C)}\ 4}.</math> | ||
− | + | ==Solution 3== | |
− | + | We can get the multiples for the numbers in the original set with multiples in the same original set | |
+ | <cmath>\begin{align*} | ||
+ | 1&: \ \text{all elements of }\{1,2,\dots,12\} \\ | ||
+ | 2&: \ 4,6,8,10,12 \\ | ||
+ | 3&: \ 6,9,12 \\ | ||
+ | 4&: \ 8,12 \\ | ||
+ | 5&: \ 10 \\ | ||
+ | 6&: \ 12 | ||
+ | \end{align*}</cmath> | ||
+ | It will be safe to start with <math>5</math> or <math>6</math> 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. | ||
− | ==Solution | + | Trying <math>4,</math> we can get <math>4,5,6,7,9,11.</math> So <math>4</math> works. |
− | We | + | Trying <math>3,</math> it won't work, so the least is <math>4.</math> This means the answer is <math>\boxed{\textbf{(C)}\ 4}.</math> |
+ | |||
+ | ==Solution 4== | ||
+ | We partition <math>\{1,2,\ldots,12\}</math> into six nonempty subsets such that for every subset, each element is a multiple of all elements less than or equal to itself: <cmath>\{1,2,4,8\}, \ \{3,6,12\}, \ \{5,10\}, \ \{7\}, \ \{9\}, \ \{11\}.</cmath> | ||
+ | Clearly, <math>S</math> must contain exactly one element from each subset: | ||
+ | |||
+ | * For <math>\{5,10\},</math> we can select either <math>5</math> or <math>10.</math> | ||
− | <math>3, | + | * For <math>\{3,6,12\},</math> we can select either <math>6</math> or <math>12.</math> Recall that since <math>9\in S,</math> we cannot select <math>3.</math> |
− | + | * For <math>\{1,2,4,8\},</math> we can select either <math>4</math> (provided that <math>12\not\in S</math>) or <math>8.</math> Recall that since either <math>6\in S</math> or <math>12\in S,</math> we can select neither <math>1</math> nor <math>2.</math> | |
− | <math>4, 5, 6, 7, 9, 11</math> | + | If <math>4\in S,</math> then the possibilities of <math>S</math> are <math>\{4,5,6,7,9,11\}</math> or <math>\{4,6,7,9,10,11\}.</math> So, the least possible value of an element in <math>S</math> is <math>\boxed{\textbf{(C)}\ 4}.</math> |
− | + | <u><b>Remark</b></u> | |
− | + | There exist multiple such partitions of <math>\{1,2,\ldots,12\}</math> into six nonempty subsets, one of which is <cmath>\{1,2,6,12\}, \ \{3,9\}, \ \{4,8\}, \ \{5,10\}, \ \{7\}, \ \{11\}.</cmath> | |
+ | Regardless of which partition we use, we will conclude that to minimize the least element of <math>S,</math> the only possibilities for <math>S</math> are <math>\{4,5,6,7,9,11\}</math> or <math>\{4,6,7,9,10,11\}.</math> | ||
− | + | ~MRENTHUSIASM | |
− | |||
− | + | ==Solution 5== | |
− | <math>2 | + | We start with 2 as 1 is not an answer option. Our set would be <math>\{2,3,5,7,11\}</math>. We realize we cannot add 12 to the set because 12 is a multiple of 3. Our set only has 5 elements, so starting with 2 won't work. |
− | <math>3 | + | We try 3 next. Our set becomes <math>\{3,4,5,7,11\}</math>. We run into the same issue as before so starting with 3 won't work. |
− | <math>4 | + | We then try 4. Our set becomes <math>\{4,5,6,7,9,11\}</math>. We see we have 6 elements with none being multiples of each other. Therefore our answer is <math>\boxed{\textbf{(C)}\ 4}.</math> |
− | + | ~[https://artofproblemsolving.com/wiki/index.php/User:South South] | |
− | + | ==Video Solution== | |
+ | https://youtu.be/M22S82Am2zM | ||
− | + | ~IceMatrix | |
− | + | ==Video Solution== | |
− | + | https://youtu.be/QtqBEZAgXpk | |
− | ~ | + | ~savannahsolver |
==See Also == | ==See Also == |
Latest revision as of 03:29, 14 October 2023
- The following problem is from both the 2018 AMC 10A #17 and 2018 AMC 12A #12, so both problems redirect to this page.
Contents
Problem
Let be a set of
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 1
We start with because
is not an answer choice. 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 valid numbers.
Finally, starting with we find that the sequence
works, giving us
Solution 2
We know that all odd numbers except namely
can be used.
Now we have possibilities to choose from for the last number (out of
). We can eliminate
and
and we have
to choose from. However,
is a multiple of
Now we have to take out either
or
from the list. If we take out
none of the numbers would work, but if we take out
we get
The least number is
so the answer is
Solution 3
We can get the multiples for the numbers in the original set with multiples in the same original set
It will be safe to start with
or
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 we can get
So
works.
Trying
it won't work, so the least is
This means the answer is
Solution 4
We partition into six nonempty subsets such that for every subset, each element is a multiple of all elements less than or equal to itself:
Clearly,
must contain exactly one element from each subset:
- For
we can select either
or
- For
we can select either
or
Recall that since
we cannot select
- For
we can select either
(provided that
) or
Recall that since either
or
we can select neither
nor
If then the possibilities of
are
or
So, the least possible value of an element in
is
Remark
There exist multiple such partitions of into six nonempty subsets, one of which is
Regardless of which partition we use, we will conclude that to minimize the least element of
the only possibilities for
are
or
~MRENTHUSIASM
Solution 5
We start with 2 as 1 is not an answer option. Our set would be . We realize we cannot add 12 to the set because 12 is a multiple of 3. Our set only has 5 elements, so starting with 2 won't work.
We try 3 next. Our set becomes . We run into the same issue as before so starting with 3 won't work.
We then try 4. Our set becomes . We see we have 6 elements with none being multiples of each other. Therefore our answer is
Video Solution
~IceMatrix
Video Solution
~savannahsolver
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.