Difference between revisions of "2014 AMC 10B Problems/Problem 24"
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We've covered all needed cases, and the two examples we found are distinct, therefore the answer is <math>\boxed{\textbf {(B) }2}</math>. | We've covered all needed cases, and the two examples we found are distinct, therefore the answer is <math>\boxed{\textbf {(B) }2}</math>. | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/45TQEV8OjRk | ||
==See Also== | ==See Also== |
Revision as of 02:13, 16 January 2020
- The following problem is from both the 2014 AMC 12B #18 and 2014 AMC 10B #24, so both problems redirect to this page.
Contents
Problem
The numbers are to be arranged in a circle. An arrangement is if it is not true that for every from to one can find a subset of the numbers that appear consecutively on the circle that sum to . Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?
Solution
We see that there are total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number is always at the top of the circle. Thus, there are only ways under rotation. Every case has exactly reflection, so that gives us only , or cases, which is not difficult to list out. We systematically list out all cases.
Now, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums and . By choosing the full circle, we can obtain . By choosing everything except for and , we can obtain subsets with sums of and .
This means that we now only need to check for and . However, once we have found a set summing to , we can choose everything else and obtain a set summing to , and similarly for and . Thus, we only need to check each case for whether or not we can obtain or .
We can make by having , or , or . We can start with the group of three. To separate from each other, they must be grouped two together and one separate, like this.
Now, we note that is next to both blank spots, so we can't have a number from one of the pairs. So since we can't have , because it is part of the pair, and we can't have there, because it's part of the pair, we must have inserted into the spot. We can insert and in and interchangeably, since reflections are considered the same.
We have and left to insert. We can't place the next to the or the next to the , so we must place next to the and next to the .
This is the only solution to make "bad."
Next we move on to , which can be made by , or , or . We do this the same way as before. We start with the three group. Since we can't have or in the top slot, we must have one there, and and are next to each other on the bottom. When we have and left to insert, we place them such that we don't have the two pairs adjacent.
This is the only solution to make "bad."
We've covered all needed cases, and the two examples we found are distinct, therefore the answer is .
Video Solution
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.