🇺🇸 AMC 8 ⇄ switch contest
1996 AMC 8 Stretch

Problem 3

Problem 3 · AMC 8 Stretch Core
Logic & Word Problems Counting & ProbabilityArithmetic & Operations account-for-all-possibilitiesorganizing-datareduce-and-expandpattern-recognition
A strip of \(15\) unit squares is cut into 'pieces.' Each piece is a run of \(1, 2, 3, 4,\) or \(5\) squares, and we use at most \(5\) pieces. List all the ways to write \(15\) as a sum of pieces under these rules (order of the pieces doesn't matter). Then connect to Gauss: one of the \(5\)-piece answers is \(5 + 4 + 3 + 2 + 1\), the numbers \(1\) through \(5\). Use this to find a quick formula for \(1 + 2 + 3 + \cdots + n\).
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Answer: 1 way with 3 pieces, 5 with 4 pieces, 12 with 5 pieces; and 1+2+…+n = n(n+1)/2
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Hint 1 of 4
Each piece is at most \(5\) squares and they must add to \(15\). What is the smallest number of pieces you could possibly use?
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Hint 2 of 4
Since \(5 + 5 + 5 = 15\), you can't do it in fewer than \(3\) pieces. So organize your search into \(3\)-piece, \(4\)-piece, and \(5\)-piece cases.
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Hint 3 of 4
In each case you're writing \(15\) as a sum of that many numbers, each between \(1\) and \(5\). Always list the biggest part first so you don't miss any or repeat any.
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Approach: Organized casework on number of pieces, then Gauss pairing
  1. The total is 15, each piece is 1 to 5 squares, at most 5 pieces. Since the biggest piece is 5 and \(5\times3=15\), you need at least 3 pieces.
  2. 3 pieces: the only way is \(5+5+5\) — 1 way.
  3. 4 pieces (biggest first): \(5+5+4+1,\ 5+5+3+2,\ 5+4+4+2,\ 5+4+3+3,\ 4+4+4+3\) — 5 ways.
  4. 5 pieces (biggest first): \(5+5+3+1+1,\ 5+5+2+2+1,\ 5+4+4+1+1,\ 5+4+3+2+1,\ 5+4+2+2+2,\ 5+3+3+3+1,\ 5+3+3+2+2,\ 4+4+4+2+1,\ 4+4+3+3+1,\ 4+4+3+2+2,\ 4+3+3+3+2,\ 3+3+3+3+3\) — 12 ways.
  5. Gauss: in \(1+2+3+4+5\), pair \(1+5=6\), \(2+4=6\), middle 3; each pair adds to \(n+1\). For \(1+2+\cdots+n\), pairing the ends always gives \(n+1\), so \(1+2+\cdots+n=\dfrac{n(n+1)}{2}\). Check: \(1+2+3+4+5=\dfrac{5\times6}{2}=15\), and \(1+2+\cdots+100=\dfrac{100\times101}{2}=5050\).
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