Difference between revisions of "2018 AMC 10A Problems/Problem 17"
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== Solution 1 == | == Solution 1 == | ||
− | If we start with <math>1</math> | + | If we start with <math>1,</math> we can include nothing else, so that won't work. (Also note that <math>1</math> is not an answer choice) |
− | If we start with <math>2</math> | + | 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 | + | 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> | + | 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== | ==Solution 2== | ||
We know that all the odd numbers except <math>1,</math> namely <math>3, 5, 7, 9, 11,</math> can be used. | We know that all the odd numbers except <math>1,</math> namely <math>3, 5, 7, 9, 11,</math> can be used. | ||
− | Now we have <math>7</math> 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> | + | Now we have <math>7</math> 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> | + | The least number is <math>4,</math> so the answer is <math>\boxed{\textbf{(C)}\ 4}.</math> |
==Solution 3== | ==Solution 3== | ||
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 | ||
− | + | <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 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. | 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 <math>4</math> | + | Trying <math>4,</math> we can get <math>4,5,6,7,9,11.</math> So <math>4</math> works. |
− | Trying <math>3</math> won't work, so the least is <math>4</math> | + | 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> |
==Video Solution== | ==Video Solution== |
Revision as of 17:47, 27 August 2021
- The following problem is from both the 2018 AMC 12A #12 and 2018 AMC 10A #17, so both problems redirect to this page.
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
If we start with we can include nothing else, so that won't work. (Also note that
is not an answer choice)
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 valid numbers.
Finally, starting with we find that the sequence
works, giving us
Solution 2
We know that all the odd numbers except namely
can be used.
Now we have 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 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 we can get
So
works.
Trying
it won't work, so the least is
This means the answer is
Video Solution
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.