Competition · AMC preparation · step 4 of 4
AMC 8 · 2022 · #20
Grade 8 arithmeticalgebra
Pick an answer.
AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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).
Find the common sum
Sum the fully filled top row to get the constant every row and column must hit: 12.
Adding signed integers like -2 to a positive sum is exactly the Grade 6 "positive and negative numbers describe quantities" skill.
6.NS.C.5Draw A DiagramName the empty cells
Name the three empty cells: a = mid-left, b = center, c = bottom-middle; the bottom-left stays x.
Naming unknown cells with letters is exactly Grade 6 "use variables to represent numbers and write expressions."
6.EE.B.6Convert To AlgebraWrite one equation per line
Set each row/column sum equal to 12, giving a + b = 13, x + c = 4, a + x = 14, and b + c = 3.
Translating each row/column constraint into an equation is Grade 6 equation writing.
6.EE.B.7Convert To AlgebraSolve everything in terms of x
Solve in terms of x: a = 14 - x, b = x - 1, c = 4 - x, and Column 2 confirms it.
Solving four linked linear equations to express a, b, c in terms of x is the Grade 8 "solve pairs of simultaneous linear equations" idea pushed to a small system.
8.EE.C.8Convert To AlgebraTurn largest into an inequality
The condition x beats a, b, c becomes x ＞ 14 - x, x ＞ x - 1, x ＞ 4 - x; the binding one is x ＞ 7.
Setting up and solving inequalities like x ＞ 14 - x is Grade 7 "construct simple equations and inequalities to solve problems."
The corner number x must be greater than 7.
▸ Why?
The tightest of the three "x beats another missing number" demands is that x beat the mid-left entry, which equals 14 - x, so the whole condition reduces to x ＞ 14 - x.
▸ Why?
The mid-left entry equals 14 - x because the three cells of column 1 are -2, the mid-left entry, and x, and they must total the shared sum 12, so -2 + (mid-left) + x = 12.
▸ Why?
The three cells of column 1 combine to make that column's single total, so their values are locked to add up to it.
▸ Why?
The shared total is 12 because the fully-filled top row -2, 9, 5 combines into its own total of 12, and every row and column is required to share that one total.
▸ Why?
The top row's three cells combine into the whole row's total, and -2 + 9 + 5 = 12.
▸ Why?
Rearranging -2 + (mid-left) + x = 12 into mid-left = 12 + 2 - x = 14 - x works because subtraction undoes the addition that built the total.
▸ Why?
The demand x ＞ 14 - x is the same as x ＞ 7: adding x to both sides gives 2x ＞ 14, and halving both sides gives x ＞ 7.
▸ Why?
Halving 2x back down to x works because dividing undoes multiplying, so 2x ＞ 14 turns into x ＞ 7.
Pick the smallest valid x
The least integer above 7 is x = 8: it gives a = 6, b = 7, c = -4 and beats all three, so (D).
Eliminating choices that violate x ＞ 7 and testing the smallest survivor is the multiple-choice version of solving an inequality for its least integer solution.
7.EE.B.4Eliminate PossibilitiesThis 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.
- Find the common sum
- Name the empty cells
- Write one equation per line
- Solve everything in terms of x
- Turn largest into an inequality
- Pick the smallest valid x
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