Difference between revisions of "2014 AMC 10B Problems/Problem 24"
m (→Video Solution) |
Sugar rush (talk | contribs) |
||
| Line 91: | Line 91: | ||
Notice how we took care of the case of 3 numbers first. That is because that case is the most severe and we usually take care of the most severe cases first in counting problems. | Notice how we took care of the case of 3 numbers first. That is because that case is the most severe and we usually take care of the most severe cases first in counting problems. | ||
| − | ==Video Solution== | + | ==Video Solution by icematrix== |
https://youtu.be/45TQEV8OjRk | https://youtu.be/45TQEV8OjRk | ||
| − | |||
| − | |||
==See Also== | ==See Also== | ||
Revision as of 22:34, 4 December 2020
- 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. 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.
![[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 
interchangeably, 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 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.
![[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 
.
Note by Williamgolly: Notice how we took care of the case of 3 numbers first. That is because that case is the most severe and we usually take care of the most severe cases first in counting problems.
Video Solution by icematrix
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. 
