2014 AMC 10B Problems/Problem 24
- The following problem is from both the 2014 AMC 12B #18 and 2014 AMC 10B #24, so both problems redirect to this page.
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, 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 find that there are only 
arrangements that satisfy these conditions. However, each of these is a reflection of another. We divide by 
for these reflections to obtain a final answer of 
.
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. 
