Problem 14 · AMC 8 Stretch
Core
Counting & Probability
pigeonholeorganizing-data
A class has 50 students. The oldest is 18 and the youngest is 15. Show that at least 2 students were born in the same month of the same year.
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Answer: at least 2 students share a birth month and year
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Hint 1 of 4
Ages from 15 to 18 cover how many different birth YEARS?
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Hint 2 of 4
Each year has 12 months. Multiply the number of years by 12 to count the possible (year, month) combinations.
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Hint 3 of 4
Ages 15–18 span 4 birth years, so there are \(4 \times 12 = 48\) possible (year, month) boxes.
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Approach: Pigeonhole — 50 students into 48 month-and-year boxes
- Students aged 15 to 18 were born in one of 4 different years.
- Each year has 12 months, so the number of possible (birth year, birth month) combinations is \(4 \times 12 = 48\). Make these 48 combinations the boxes.
- Put each of the 50 students into the box for their birth month and year. Since \(50 > 48\), some box holds at least 2 students.
- Those 2 students were born in the same month of the same year.
Mark:
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