Problem 5 · AMC 8 Stretch
Stretch
Number Theory
Counting & Probability
pigeonholeorganizing-data
Pick any 27 different odd numbers, each less than 100. Show that two of them must add up to 102.
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Answer: two of them sum to 102
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Hint 1 of 4
First, how many odd numbers are below 100? They are \(1, 3, 5, \dots, 99\).
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Hint 2 of 4
Pair them up so each pair adds to 102: \(\{3,99\}, \{5,97\}, \dots, \{49,53\}\). Which numbers are left over with no partner?
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Hint 3 of 4
The leftovers are 1 (its partner 101 is too big) and 51 (its partner would be 51 again). Put those two alone in their own boxes. Count all the boxes.
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Approach: Pigeonhole — group odds below 100 into pairs summing to 102
- The odd numbers below 100 are \(1, 3, 5, \dots, 99\) — that's 50 numbers.
- Group them into boxes whose two numbers add to 102: \(\{3,99\}, \{5,97\}, \{7,95\}, \dots, \{49,53\}\). That's 24 pairs.
- Two numbers have no partner: 1 (since \(102-1=101\) isn't under 100) and 51 (since \(102-51=51\) is itself). Give each its own box. Total boxes: \(24 + 2 = 26\).
- Choose your 27 numbers and drop them in. Since \(27 > 26\), some box gets 2 numbers. A singleton box holds only 1, so the doubled box must be one of the 24 pairs — and that pair sums to \(102\).
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