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.
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 1
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 
.
Solution 2
Like Solution 1, we note that the numbers from 1-5 are always going to be able to be made. Also, by selecting all numbers but one of 
, we can obtain the numbers from 10-14, and 15 will always be made from all the numbers. So we must check only 6-9. But if one circle can make a 6, we can select all of the other numbers and get a 9. Similarly, we can do the same for 7 and 8. So we must check only for 6 and 7.
We can make by having 
, or 
, or 
. We can start with the group of three. To seperate from eachother, they must be grouped two together and one seperate, like this.
![[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$x$", A, N); label("$y$", C, SW); label("$z$", D, SE); [/asy]](http://latex.artofproblemsolving.com/f/a/6/fa6111802ffd528eeb2c06d1c373e3f504cfdf98.png)
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 
interchangably, since reflections are considered the same.
![[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$3$", A, N); label("$2$", C, SW); label("$1$", D, SE); [/asy]](http://latex.artofproblemsolving.com/1/e/d/1ed6f2120054fc5856a5dc8e90e93a6894261f3b.png)
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 .
![[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$3$", A, N); label("$5$", B, NW); label("$2$", C, SW); label("$1$", D, SE); label("$4$", E, NE); [/asy]](http://latex.artofproblemsolving.com/e/2/6/e263acfc14751b36119aa9fe51f58e6c33dc1785.png)
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 4 or 2 in the top slot, we must have one there, and 4 and 2 are next to eachother on the bottom. When we have and left to insert, we place them such that we don't have the two pairs adjacent.
![[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$1$", A, N); label("$3$", B, NW); label("$2$", C, SW); label("$4$", D, SE); label("$5$", E, NE); [/asy]](http://latex.artofproblemsolving.com/f/1/7/f1704690dde3a44eeb4765b253838b226edfea06.png)
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 .
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. 
