Problem 10 · 1990 AJHSME
Hard
Algebra & Patterns
calendar-arithmeticsubstitution
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Answer: A — P.
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Hint 1 of 2
You don't need the real dates — only how the squares relate. On any calendar, the box just below a date is +7 (a week later), and the box to the right is +1 (the next day). Write every letter as 'C plus something.'
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Hint 2 of 2
Express each letter as an offset from C, add the ones you're told to, and just read off which letter matches the total. The actual month never matters — pick C = 0 if you like.
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Approach: label every box as an offset from C (down = +7, right = +1)
- Anchor on C. Moving one box right adds 1 (next day); moving one box down adds 7 (next week). Reading the grid: A is the box right of C, so A = C + 1. B sits two rows below and one column left of C, so B = C + 14 − 1 = C + 13.
- We need a letter X with X + C = A + B. The right side is (C+1) + (C+13) = 2C + 14, so X = C + 14.
- C + 14 is two rows straight down from C — that box is P. (Quick check with a real month: if C = 8, A = 9, B = 21, P = 22, and 22 + 8 = 30 = 9 + 21. ✓)
- *Why this transfers:* on calendar-grid puzzles, never plug in real dates — turn each cell into 'anchor ± (7×rows + 1×columns)' and the arithmetic collapses.
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