Problem 20 · 2010 AMC 8
Hard
Counting & Probability
inclusion-exclusiondivisibility
In a room, 2/5 of the people are wearing gloves, and 3/4 of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?
Show answer
Answer: A — 3.
Show hints
Hint 1 of 3
‘2/5 of the people’ has to be a whole number, and so does ‘3/4 of the people.’ That forces the head count to be a multiple of both 5 and 4 — pick the smallest such number.
Still stuck? Show hint 2 →
Hint 2 of 3
To minimize overlap, push the two groups as far apart as you can. But gloves (8) + hats (15) = 23 people-slots in a room of only 20 — the 3 extra slots must double up. That forced overlap is the minimum.
Still stuck? Show hint 3 →
Hint 3 of 3
This is inclusion-exclusion read as a floor: min(both) = gloves + hats − total, whenever that's positive.
Show solution
Approach: smallest legal room, then the forced overlap
- Both 2/5 and 3/4 of the people must be whole numbers, so the total is a multiple of 5 and 4 → a multiple of 20. The smallest room is 20 people (smaller total gives fractional people).
- Then gloves = 2/5 · 20 = 8 and hats = 3/4 · 20 = 15.
- Those are 8 + 15 = 23 ‘wearings’ spread over 20 people. Even spreading them out as much as possible, 23 − 20 = 3 of them are forced to land on someone already counted. So at least 3 wear both.
- Why this transfers: ‘A + B exceeds the whole’ guarantees an overlap of at least (A + B − whole) — a pigeonhole-flavored floor that shows up whenever two big groups share a small population.
Mark:
· log in to save