2024 AMC 10A Problems/Problem 25
- The following problem is from both the 2024 AMC 10A #25 and 2024 AMC 12A #22, so both problems redirect to this page.
Problem
The figure below shows a dotted grid 
cells wide and 
cells tall consisting of 
squares. Carl places 
-inch toothpicks along some of the sides of the squares to create a closed loop that does not intersect itself. The numbers in the cells indicate the number of sides of that square that are to be covered by toothpicks, and any number of toothpicks are allowed if no number is written. In how many ways can Carl place the toothpicks?
![[asy] size(6cm); for (int i=0; i<9; ++i) { draw((i,0)--(i,3),dotted); } for (int i=0; i<4; ++i){ draw((0,i)--(8,i),dotted); } for (int i=0; i<8; ++i) { for (int j=0; j<3; ++j) { if (j==1) { label("1",(i+0.5,1.5)); }}} [/asy]](http://latex.artofproblemsolving.com/9/d/1/9d176dfd41e3e8e875c6b5d84fdb4f5e4f7d43a6.png)
(Best) Solution 1
Observations:
1. You can not have a vertical line in any place other than the first and second columns and the last and second-to-last columns.
2. You can place a box around the top row or along the bottom row, otherwise all the solutions have a vertical line at the first and second columns and the last and second-to-last columns.
Thus, Using casework, we can split this problem into 4 cases. However, we can focus on only the first one for right now.
For case 1, we assume that the green lines shown below are given (always have toothpicks on them)
And the only toothpicks we can place that will connect to the red lines are to go horizontally inward:
Now, concentrate on the first row of squares. A toothpick can be placed on either the bottom or top and connected to a continuous squiggle by adding vertical toothpicks, for example,
How many squiggles are possible?
We can summarize this by giving a high squiggle position a 1 and a low position a 0, thus we have a 6-digit binary sequence. Thus, we can have 
ways to make this squiggle.
Case 2: We can also, as the observations state, pull in one of the sides, thus we can have a squiggle with 5 binary digits. only utilizing the first 7 columns:
Here, we only have 5 binary digits to work with, so there are 
ways to make this squiggle.
Case 3: Similarly, we can utilize the last 7 columns.
Again, we only have 5 binary digits to work with, so there are ways to make this squiggle.
Case 4: We can use an even smaller section. Using only the middle 6 columns gives us a 4-wide squiggle:
Thus, there are 
ways to make this squiggle.
Adding up all our cases: 
However, there are two more ways to draw a qualifying shape:
We can draw a rectangle like that in the first row or third row. Thus, we have a grand total of
ways.
A note to (potential) editors:
This answer was not made to be concise or especially professional. It was made to explicitly explain this problem in a way so that it is easy to understand and follow.
~hermanboxcar5
Solution 2 (Cheese)
Notice that for any case where the closed loop does not connect from the top side of the ones and bottom side of the ones, there are two of these cases. A cheese solution can be found from this; noting that B and C are the only two options to each other, and, being two apart, with people likely to forget this case, 
is likely to be the correct answer.
Cheese solution done by juwushu.
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 | ||
| 2024 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 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. 
