Difference between revisions of "2025 AMC 8 Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
| − | First, we note that there are <math>18</math> "pairs" of squares folded on top of each other after the folding. The minimum number of gold pairs occurs when silver squares occupy the maximum number of pairs, and the maximum number of gold pairs occurs when silver squares occupy the minimum number of pairs. The former case occurs when all <math>13</math> silver squares are placed in different pairs, resulting in <math>18-13=5</math> gold pairs. The latter case occurs when the silver squares are paired up as much as possible, resulting in <math>6</math> complete pairs and another square occupying another pair slot. Then there are <math>18-7=11</math> gold pairs. Our answer is <math>11+5=\boxed{\textbf{( | + | First, we note that there are <math>18</math> "pairs" of squares folded on top of each other after the folding. The minimum number of gold pairs occurs when silver squares occupy the maximum number of pairs, and the maximum number of gold pairs occurs when silver squares occupy the minimum number of pairs. The former case occurs when all <math>13</math> silver squares are placed in different pairs, resulting in <math>18-13=5</math> gold pairs. The latter case occurs when the silver squares are paired up as much as possible, resulting in <math>6</math> complete pairs and another square occupying another pair slot. Then there are <math>18-7=11</math> gold pairs. Our answer is <math>11+5=\boxed{\textbf{(C16} </math>. |
| − | == | + | ==Solution 2== |
| + | We can see that the least number of gold-on-gold pairs will be obtained when the 13 silver squares are placed on the two sides so that they don't overlap when folded over (because then it will minimize the number of gold-on-golds). We can see that if we split them up 6 and 7 on both sides, and then fold it, the number of gold-on-golds will be 18-13 = 5. | ||
| + | |||
| + | The maximum number of gold-on-golds will be achieved when the silver squares overlap when folded over, which will increase the number of gold-on-golds. If we align 6 silver squares with each other on each side, and put the last one somewhere else, we get the maximum is 18-7 = 11. Therefore, the answer is <math>11+5=\boxed{\textbf{(C)}~16}</math>. | ||
| + | |||
| + | ~Sigmacuber | ||
| + | |||
| + | ==Video Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| + | You can contact him through his Youtube Channel | ||
| + | |||
| + | ==Video Solution by Thinking Feet== | ||
| + | https://youtu.be/PKMpTS6b988 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2025|num-b=14|num-a=16}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:36, 31 January 2025
Contents
Problem
Kei draws a 
-by- grid. He colors 
of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let 
and 
equal equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of 
?
Solution 1
First, we note that there are 
"pairs" of squares folded on top of each other after the folding. The minimum number of gold pairs occurs when silver squares occupy the maximum number of pairs, and the maximum number of gold pairs occurs when silver squares occupy the minimum number of pairs. The former case occurs when all silver squares are placed in different pairs, resulting in 
gold pairs. The latter case occurs when the silver squares are paired up as much as possible, resulting in complete pairs and another square occupying another pair slot. Then there are 
gold pairs. Our answer is $11+5=\boxed{\textbf{(C16}$ (Error compiling LaTeX. Unknown error_msg).
Solution 2
We can see that the least number of gold-on-gold pairs will be obtained when the 13 silver squares are placed on the two sides so that they don't overlap when folded over (because then it will minimize the number of gold-on-golds). We can see that if we split them up 6 and 7 on both sides, and then fold it, the number of gold-on-golds will be 18-13 = 5.
The maximum number of gold-on-golds will be achieved when the silver squares overlap when folded over, which will increase the number of gold-on-golds. If we align 6 silver squares with each other on each side, and put the last one somewhere else, we get the maximum is 18-7 = 11. Therefore, the answer is 
.
~Sigmacuber
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI You can contact him through his Youtube Channel
Video Solution by Thinking Feet
See Also
| 2025 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
