Problem 18 · AMC 8 Stretch
Core
Number Theory
Counting & Probability
pigeonholeorganizing-data
Pick any 7 numbers from \(1, 2, 3, \dots, 12\). Show that two of them differ by exactly 6.
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Answer: two of them differ by 6
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Hint 1 of 4
You want two numbers that differ by 6. Could you sort the numbers 1 to 12 into groups so that 'same group' automatically means 'differ by 6'?
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Hint 2 of 4
Pair the numbers so each pair differs by 6: \(\{1,7\}, \{2,8\}, \{3,9\}, \{4,10\}, \{5,11\}, \{6,12\}\). How many pairs is that, and do they use up all 12 numbers?
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Hint 3 of 4
There are 6 pairs covering all 12 numbers — your 6 boxes. You're picking 7 numbers.
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Approach: Pigeonhole — pair the numbers into 6 difference-6 boxes
- Split \(1, 2, \dots, 12\) into pairs that differ by 6: \(\{1,7\}, \{2,8\}, \{3,9\}, \{4,10\}, \{5,11\}, \{6,12\}\).
- That's 6 pairs, and together they use every number from 1 to 12. Make these 6 pairs the boxes.
- Pick your 7 numbers and drop them into the boxes. Since \(7 > 6\), two of them land in the same pair.
- The two numbers in a pair differ by exactly \(6\).
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