2018 AMC 10A Problems/Problem 17
Contents
Problem
Let be a set of 6 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
If we start with , we can include nothing else, so that won't work.
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 nowhere else to go.
Finally, starting with , we find that the sequence
works, giving us
.
(Random_Guy)
Solution 2
We know that all the odd numbers can be used.
Now we have 7 to choose from for the last number (out of ). We can eliminate 1, 2, 10, and 12, and we have
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
, none of the numbers would work, but if we take out
, we get:
So the least number is , so the answer is
.
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 |
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