2022 AIME II Problems/Problem 1
Contents
Problem
Adults made up of the crowd of people at a concert. After a bus carrying more people arrived, adults made up of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.
Solution 1
Let be the number of people at the party before the bus arrives. We know that , as of people at the party before the bus arrives are adults. Similarly, we know that , as of the people at the party are adults after the bus arrives. can be reduced to , and since we are looking for the minimum amount of people, is . That means there are people at the party after the bus arrives, and thus there are adults at the party.
~eamo
Solution 2 (Kind of lame)
Since at the beginning, adults make up of the concert, the amount of people must be a multiple of 12.
Call the amount of people in the beginning .Then must be divisible by 12, in other words: must be a multiple of 12. Since after 50 more people arrived, adults make up of the concert, is a multiple of 25. This means must be a multiple of 5.
Notice that if a number is divisible by 5, it must end with a 0 or 5. Since 5 is impossible (obviously, since multiples of 12 end in 2, 4, 6, 8, 0,...), must end in 0.
Notice that the multiples of 12 that end in 0 are: 60, 120, 180, etc.. By trying out, you can clearly see that is the minimum number of people at the concert.
So therefore, after 50 more people arrive, there are people at the concert, and the number of adults is . Therefore the answer is .
I know this solution is kind of lame, but this is still pretty straightforward. This solution is very similar to the first one, though.
~hastapasta
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=gBIxZ6SUr_w
Video Solution by Power of Logic
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.