AMC 8 · 2022 · #20

Grade 8 arithmeticalgebra
Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration
systems-of-equationslinear-equations-one-varlogical-deduction convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: linear-equations-one-varmulti-digit-arithmetic
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A 3 × 3 grid is filled with integers so that every row and every column has the same sum. Five entries are already filled: top row is -2, 9, 5; the middle-right cell is -1; the bottom-right cell is 8; and the bottom-left cell is x. Four cells are missing. The number x in the lower-left corner must be strictly larger than each of the other three missing numbers. Find the smallest possible integer value of x.

Pick an answer.

(A)
-1
(B)
5
(C)
6
(D)
8
(E)
9

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Convert to Algebra

The top row is fully filled, so we instantly read off the magic sum -2 + 9 + 5 = 12. After that, each missing cell sits in a row or column with only one unknown, so naming the missing cells a, b, c and writing row sum = 12 / column sum = 12 turns the puzzle into a tiny system of linear equations — Tool #13 (Algebra) is the cleanest path. Tool #1 (Diagram) is the labelled 3 × 3 grid we keep in front of us so we don't lose track of which cell is which. Once we express a, b, c in terms of x, the condition "x beats the others" becomes three simple inequalities, and Tool #3 (Eliminate) lets us check the answer choices {-1, 5, 6, 8, 9} against x ＞ 7 to land on (D).

1STEP 1

Sum the fully filled top row to get the constant every row and column must hit: 12.

-2 + 9 + 5 = 12
2STEP 2

Name the three empty cells: a = mid-left, b = center, c = bottom-middle; the bottom-left stays x.

a = mid-left, b = center, c = bottom-middle
3STEP 3

Set each row/column sum equal to 12, giving a + b = 13, x + c = 4, a + x = 14, and b + c = 3.

a + b = 13, x + c = 4, a + x = 14, b + c = 3
4STEP 4

Solve in terms of x: a = 14 - x, b = x - 1, c = 4 - x, and Column 2 confirms it.

a = 14 - x, b = x - 1, c = 4 - x
5STEP 5

The condition x beats a, b, c becomes x ＞ 14 - x, x ＞ x - 1, x ＞ 4 - x; the binding one is x ＞ 7.

x ＞ 7
6STEP 6

The least integer above 7 is x = 8: it gives a = 6, b = 7, c = -4 and beats all three, so (D).

x = 8 → a = 6, b = 7, c = -4 → (D)
Answer
8
Plug x = 8 back into the full grid: row 1 = -2 + 9 + 5 = 12, row 2 = 6 + 7 + (-1) = 12, row 3 = 8 + (-4) + 8 = 12; columns are -2 + 6 + 8 = 12, 9 + 7 + (-4) = 12, 5 + (-1) + 8 = 12. All six sums match, and 8 is strictly greater than 6, 7, -4. The smaller answer choices -1, 5, 6 all fail x ＞ 7, so x = 8 is genuinely the minimum.
💡Key takeaway

This AMC 8 problem only needs Grade 8 "solve a small system of linear equations" you already know — once the magic sum 12 is in hand, the four cells fall out as expressions in x and the inequality x ＞ 7 gives x = 8.

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