🇺🇸 AMC 8 ⇄ switch contest
2018 AMC 8

Problem 19

Problem 19 · 2018 AMC 8 Hard
Counting & Probability caseworkcareful-counting
Figure for AMC 8 2018 Problem 19
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Answer: C — 8 ways.
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Hint 1 of 2
Instead of testing all 16 bottom rows, notice the pyramid is reversible: if you know a cell and the cell to its bottom-left, the bottom-right cell is forced. So a whole row plus one free corner choice determines the row beneath it.
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Hint 2 of 2
The technique is counting free choices: building downward, each new row is pinned down by the row above plus exactly one free sign at the left. Three rows get built below the fixed top, so the count is 2 × 2 × 2.
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Approach: count the free choices building downward
  1. Key reversibility: given a cell's sign and its bottom-left neighbor, the rule forces the bottom-right neighbor (if the top is +, the two bottoms match; if −, they differ). So a known row plus one free leftmost choice determines the entire row below it.
  2. Start from the fixed + at the apex and build down. Row 2 (2 cells): pick its left cell freely (2 ways), the right is forced — 2 choices. Row 3 is then determined by row 2 plus one free left cell — another 2. Row 4 (the bottom) similarly — another 2.
  3. Total = 2 × 2 × 2 = 8.
  4. You'll see it again: when a structure is "reversible with one free knob per level," the answer is 2 to the power of the number of free knobs — far faster than brute-forcing all configurations.
Another way — XOR / parity of the bottom row:
  1. Encode + as 0 and − as 1; then "+ when the two below match" is exactly addition mod 2 (XOR): each cell is the XOR of the two beneath it.
  2. Stacking three layers, the top equals bottom1 ⊕ bottom2 ⊕ bottom3 ⊕ bottom4, because the weights are the row-3 binomial coefficients 1, 3, 3, 1 — all odd, so all four corners count.
  3. Top = + means an even number of −'s in the bottom row. Of the 16 bottom rows, exactly half have even parity: 8.
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