Problem 24 · 1993 AJHSME
Stretch
Number Theory
patternrows
The figure below shows a triangular ‘staircase’ array of numbers. The first row has 1 number, the second row has 3, the third row has 5, and so on (the kth row has 2k−1 numbers, in order).
1
2 3 4
5 6 7 8 9
10 11 12 13 14 15 16
…
What number is directly above 142 in this array of numbers?
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Answer: C — 120.
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Hint 1 of 2
The right edge of each row is the giveaway: rows end at 1, 4, 9, 16, … — the perfect squares k². So spotting which square 142 sits just under tells you its row instantly.
Still stuck? Show hint 2 →
Hint 2 of 2
Each row is centered and 2 numbers wider than the one above (one extra on each side). So the entry above isn't in the same position — it's shifted one slot toward the left.
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Approach: locate by row, then account for the centering shift
- Rows end at the squares 1, 4, 9, …, k². Since 11² = 121 and 12² = 144, the number 142 lives in row 12 (which runs 122 to 144). Its position in that row is 142 − 122 + 1 = 21st, out of 12's 23 entries.
- Row 11 runs 101 to 121 — 21 entries — sitting centered above row 12. Because row 12 has one extra number on the left, row 12's 21st entry lines up under row 11's 20th entry.
- Row 11's 20th entry is 101 + (20 − 1) = 120.
- Why this transfers: two facts crack any centered number-triangle — the row-ending rule (here, squares) pins the row, and 'each row is 1 wider on each side' gives the alignment shift. Forgetting the shift is the classic slip; 142 is 21st, but directly above is the 20th, not the 21st.
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