🇺🇸 AMC 8 ⇄ switch contest
1996 AMC 8 Stretch

Problem 4

Problem 4 · AMC 8 Stretch Core
Counting & Probability Geometry & Measurement reduce-and-expandpattern-recognitionorganizing-data
How many squares of all sizes are there on an \(8 \times 8\) checkerboard?
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Answer: 204 squares
Show hints
Hint 1 of 4
Careful — the answer is NOT just 64! Those are only the small 1×1 squares. There are also 2×2 squares, 3×3 squares, all the way up to the whole 8×8 board. Start with a tiny board to get the idea.
Still stuck? Show hint 2 →
Hint 2 of 4
Count all the squares on a 1×1 board, then a 2×2 board, then a 3×3. For example, a 2×2 board has four 1×1 squares plus one 2×2 square = 5 total. You should get totals 1, 5, 14, ...
Still stuck? Show hint 3 →
Hint 3 of 4
On the 8×8 board, count each size separately. A \(k\)-by-\(k\) square can slide into \((9-k)\) positions across and \((9-k)\) down, so there are \((9-k)^2\) of them. That gives \(8^2 + 7^2 + \dots + 1^2\).
Show solution
Approach: Reduce and expand — sum the per-size counts \((9-k)^2\)
  1. There are squares of every size from 1×1 up to 8×8, not just the 64 unit squares.
  2. How many \(k\)-by-\(k\) squares fit? Its left edge can start in any of \((9-k)\) columns and its top edge in any of \((9-k)\) rows, so there are \((9-k)^2\) of size \(k\):
  3. Size \(k\)Count \((9-k)^2\)
    164
    249
    336
    425
    516
    69
    74
    81
  4. Add them all up: \(64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204\). This is \(1^2 + 2^2 + \dots + 8^2\); the formula \(\dfrac{n(n+1)(2n+1)}{6}\) for \(n = 8\) gives \(\dfrac{8 \times 9 \times 17}{6} = 204\) too.
  5. So the board holds \(204\) squares of all sizes.
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