Problem 7 · 1985 AJHSME
Medium
Algebra & Patterns
pattern-by-row
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Answer: C — 36.
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Hint 1 of 2
Don't picture all 37 rows — just look at how one row is built. Each row starts AND ends with white, with blacks tucked between. If the whites are like fence posts, the blacks are the gaps between them. How many gaps for a given number of posts?
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Hint 2 of 2
Each row reads white-black-white-black…-white. With the blacks sandwiched strictly between whites, there's always exactly one fewer black than white. Find the white count by the row number, subtract 1.
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Approach: find the row-n pattern
- Count the rows shown: row 1 is 1 white, 0 black; row 2 is W B W (2 white, 1 black); row 3 is W B W B W (3 white, 2 black). Each row's white count equals its row number, and the blacks fill the gaps between whites — one fewer.
- So row n has (n − 1) black squares. Row 37: 37 − 1 = 36.
- Why this transfers: 'posts and gaps' shows up everywhere — n fence posts make n − 1 gaps, n trees in a row leave n − 1 spaces. Whenever items strictly alternate and the ends match, the inner kind is one fewer.
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