2022 AIME II Problem 1

Below is the professionally curated solution for Problem 1 of the 2022 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AIME II solutions, or check the answer key.

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Concepts:fractiondivisibilityoptimization

Difficulty rating: 1890

1.

Adults made up 512\frac{5}{12} of the crowd of people at a concert. After a bus carrying 5050 more people arrived, adults made up 1125\frac{11}{25} of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.

Solution:

Let the original crowd have 12k12k people, of whom 5k5k are adults. After the bus arrives there are 12k+5012k + 50 people, and the number of adults is 1125(12k+50).\frac{11}{25}(12k + 50). For this to be an integer, 2525 must divide 12k+50,12k + 50, so 2512k,25 \mid 12k, and since gcd(12,25)=1\gcd(12, 25) = 1 this means kk is a multiple of 25.25.

The adult count 1125(12k+50)\frac{11}{25}(12k + 50) increases with k,k, so the minimum occurs at k=25:k = 25: the new total is 350350 and the number of adults is 1125350=154.\frac{11}{25} \cdot 350 = 154. This is achievable, for example if the bus carries 2929 adults and 2121 non-adults, so the answer is 154.154.

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