Difference between revisions of "2018 AMC 10A Problems/Problem 17"
m (→Solution) |
(→Solution 2) |
||
Line 26: | Line 26: | ||
-Baolan | -Baolan | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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>2: 4,6,8,10,12</math> | ||
+ | |||
+ | <math>3: 6,9,12</math> | ||
+ | |||
+ | <math>4: 8,12</math> | ||
+ | |||
+ | <math>5: 10</math> | ||
+ | |||
+ | <math>6: 12</math> | ||
+ | |||
+ | 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>, 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>. This means the answer is <math>\boxed{\textbf{(C) } 4}</math> | ||
+ | |||
+ | ~OlutosinNGA | ||
==See Also == | ==See Also == |
Revision as of 10:20, 24 January 2019
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 no more places to go.
Finally, starting with , we find that the sequence works, giving us . (Random_Guy)
Solution 2
We know that all the odd numbers (except 1) 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 .
-Baolan
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 won't work, so the least is . This means the answer is
~OlutosinNGA
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.