Difference between revisions of "2018 AMC 10A Problems/Problem 17"
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* For <math>\{5,10\},</math> we can select either <math>5</math> or <math>10.</math> | * For <math>\{5,10\},</math> we can select either <math>5</math> or <math>10.</math> | ||
− | * For <math>\{3,6,12\},</math> we can select either <math>6</math> or <math>12.</math> Recall that | + | * For <math>\{3,6,12\},</math> we can select either <math>6</math> or <math>12.</math> Recall that as <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 | + | * 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 as either <math>6\in S</math> or <math>12\in S,</math> we can select neither <math>1</math> nor <math>2.</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> | 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> |
Revision as of 01:21, 5 January 2022
- 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. By the way, 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 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 as
we cannot select
- For
we can select either
(provided that
) or
Recall that as 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
Video Solution
~IceMatrix
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.