Problem 8 · AMC 8 Stretch
Stretch
Number Theory
Counting & ProbabilityAlgebra & Patterns
pattern-recognitionintelligent-guessing-and-testingorganizing-data
Bending a wire of whole-number length \(n\) at two marks to make a triangle gives some number of bending-point choices. The counts for lengths 3 through 15 are: 1, 0, 3, 1, 6, 3, 10, 6, 15, 10, 21, 15, 28. The row looks scrambled. What sequence of numbers is hidden inside it? (Hint: look at odd lengths and even lengths separately.)
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Answer: the triangular numbers 1, 3, 6, 10, 15, 21, 28, ...
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Hint 1 of 4
The numbers jump around because TWO patterns are tangled together. Separate them: write down only the odd-length answers, then only the even-length answers.
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Hint 2 of 4
Odd lengths 3,5,7,9,11,13,15 give 1,3,6,10,15,21,28. Even lengths 4,6,8,10,12,14 give 0,1,3,6,10,15. Do you recognize 1,3,6,10,15,...?
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Hint 3 of 4
These are the TRIANGULAR NUMBERS: 1, 3, 6, 10, 15, 21, 28, ... formed by 1, 1+2, 1+2+3, 1+2+3+4, and so on. The \(k\)-th one is \(1+2+\cdots+k\).
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Approach: Split odd and even lengths to reveal triangular numbers
- Separate the row by parity. Odd lengths 3, 5, 7, 9, 11, 13, 15 give 1, 3, 6, 10, 15, 21, 28. Even lengths 4, 6, 8, 10, 12, 14 give 0, 1, 3, 6, 10, 15.
- Both lists are the triangular numbers \(T_k = 1 + 2 + \cdots + k\): \(T_1=1, T_2=3, T_3=6, T_4=10, T_5=15, T_6=21, T_7=28\).
- Why? Counting by the first bend gives a staircase \(1 + 2 + 3 + \cdots\), exactly the X-staircase from the 9-inch wire. Odd lengths start one row higher than even lengths (an even wire wastes its exact-middle mark), so the even list is shifted down.
- So the hidden pattern is the triangular numbers \(1, 3, 6, 10, 15, 21, 28, \dots\), interleaved — and an even-length wire gives the same count as the odd-length wire 3 inches shorter (14 gives 15, same as 11; 10 gives 6, same as 7).
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