Difference between revisions of "2024 AIME I Problems/Problem 3"
| Line 47: | Line 47: | ||
<math>15</math> coins: <math>L</math> | <math>15</math> coins: <math>L</math> | ||
| − | We can see that losing positions occur when <math>n</math> is congruent to <math>0</ | + | We can see that losing positions occur when <math>n</math> is congruent to <math>0, </math>2 \mod{5})<math> and winning positions occur otherwise. As </math>n<math> ranges from </math>1<math> to </math>2020<math>, </math>\frac{2}{5}<math> of these values are losing positions where Bob will win. As </math>n<math> ranges from </math>2021<math> to </math>2024<math>, </math>2022<math> is the only value where Bob wins. Therefore, the answer is </math>2020\times\frac{2}{5}+1=\boxed{809}$ |
Revision as of 19:28, 2 February 2024
Contents
Problem
Alice and Bob play the following game. A stack of 
tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either 
token or 
tokens from the stack. Whoever removes the last token wins. Find the number of positive integers less than or equal to 
for which there exists a strategy for Bob that guarantees that Bob will win the game regardless of Alice's play.
Solution 1
Let's first try some experimentation. Alice obviously wins if there is one coin. She will just take it and win. If there are 2 remaining, then Alice will take one and then Bob will take one, so Bob wins. If there are 
, Alice will take , Bob will take one, and Alice will take the final one. If there are , Alice will just remove all at once. If there are 
, no matter what Alice does, Bob can take the final coins in one try. Notice that Alice wins if there are , , or coins left. Bob wins if there are 
or coins left.
After some thought, you may realize that there is a strategy for Bob. If there is n is a multiple of , then Bob will win. The reason for this is the following: Let's say there are a multiple of coins remaining in the stack. If Alice takes , Bob will take , and there will still be a multiple of . If Alice takes , Bob will take , and there will still be a multiple of . This process will continue until you get 
coins left. For example, let's say there are 
coins. No matter what Alice does, Bob can simply just do the complement. After each of them make a turn, there will always be a multiple of left. This will continue until there are coins left, and Bob will end up winning.
After some more experimentation, you'll realize that any number that is congruent to mod will also work. This is because Bob can do the same strategy, and when there are coins left, Alice is forced to take and Bob takes the final coin. For example, let's say there are 
coins. If Alice takes , Bob will take . If Alice takes , Bob will take . So after they each make a turn, the number will always be equal to mod . Eventually, there will be only coins remaining, and we've established that Alice will simply take and Bob will take the final coin.
So we have to find the number of numbers less than or equal to that are either congruent to mod or mod . There are 
numbers in the first category: 
. For the second category, there are 
numbers. 
. So the answer is 
~lprado
Solution 2
We will use winning and losing positions.
coin: 
coins: 
coins:
coins:
coins:
coin:
coins:
coins:
coins:
coins:
coin:
coins:
coins:
coins:
coins:
We can see that losing positions occur when is congruent to 
2 \mod{5})
n
1
2020
\frac{2}{5}
n202120242022
2020\times\frac{2}{5}+1=\boxed{809}$
See also
| 2024 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
