Problem 20 · 2022 AMC 8
Hard
Algebra & Patterns
substitutionsum-constraint
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Answer: D — x = 8.
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Hint 1 of 2
One row is completely filled in — that quietly tells you the shared sum, which unlocks every blank.
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Hint 2 of 2
The common sum is −2 + 9 + 5 = 12. Use that to write each of the three other missing cells in terms of x; then “x is the largest” turns into a few inequalities.
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Approach: the full top row gives the common sum; express every blank in terms of x
- Insight: the top row is fully known, so it hands you the magic sum for free: −2 + 9 + 5 = 12. Every row and column must total 12, so each blank is forced once you write it against that 12.
- Cell above x (first column): −2 + ? + x = 12 ⇒ ? = 14 − x.
- Middle-bottom (bottom row): x + ? + 8 = 12 ⇒ ? = 4 − x.
- Center (middle row): (14−x) + ? + (−1) = 12 ⇒ ? = x − 1.
- Now “x beats all three”: x > 14−x ⇒ x > 7 (the binding one); x > 4−x and x > x−1 are easier. Smallest integer with x > 7 is 8.
- Sanity check at x = 8: blanks become 6, −4, 7; the grid reads [−2, 9, 5 / 6, 7, −1 / 8, −4, 8] — every row and column sums to 12, and 8 is the largest of {6, −4, 7, 8}. ✓
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