AMC 8 · 1999 · #11

Grade 6 arithmetic

Archetype Pair-Sum Invariant └ Sub-archetype Uniform Pair-Sum Recovery

pattern-recognitionset-partitionsystematic-enumerationsequences-arithmetic identify-subproblemspattern-recognition ↑ Prerequisites: multi-digit-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Place the five numbers 1, 4, 7, 10, 13 in the five squares of a plus-sign shape so that the three numbers in the horizontal row add up to the same value as the three numbers in the vertical column. Find the largest possible value of that common sum.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Find an Invariant

Drawing the plus sign (Tool #1) makes the key feature visible: the center square sits inside the row AND inside the column. If you add the row sum and the column sum, every outside number is counted once but the center number is counted twice. That gives the invariant equation 2S = 35 + center — Tool #11 (Find an Invariant). Since 35 is fixed, the equation pins S entirely to the center value, so maximizing S just means making the center as large as possible. No guessing of arm placements is needed.

1STEP 1

Sketch the plus sign: the row (left, center, right) and the column (top, center, bottom) share the center square.

shared square = center

2STEP 2

Add row + column: the four arm numbers count once, the center counts twice, and all five total 35.

S + S = (1 + 4 + 7 + 10 + 13) + center = 35 + center

3STEP 3

Solve the invariant equation for S. Divide both sides by 2.

2S = 35 + center → S =

\frac{35 + center}{2}

4STEP 4

Bigger center means bigger S, so put the largest number 13 in the center: S = $\frac{35 + 13}{2}$ = 24 (48 is even, so S is whole).

center = 13 → S =

\frac{35 + 13}{2}

\frac{48}{2}

= 24

5STEP 5

Split the rest {1, 4, 7, 10} into pairs summing 11: (1, 10) and (4, 7). Then row 1+13+10 and column 4+13+7 both equal 24 → (D).

row: 1 + 13 + 10 = 24, column: 4 + 13 + 7 = 24 → (D)

Answer

Check the other choices against the invariant S =

\frac{35 + center}{2}

. Center = 1 → S = 18; center = 7 → S = 21; center = 13 → S = 24. Centers 4 and 10 make 35 + center odd, so S isn't a whole number and those centers don't work. So the only achievable common sums are 18, 21, 24, and the maximum 24 matches answer (D). Choice (E) 30 exceeds

\frac{35 + 13}{2}

= 24, so no center makes it reachable — confirming (D).

💡Key takeaway

The center square is counted twice when you add the row and the column, so 2S = 35 + center. To make S biggest, put the biggest number (13) in the center — that gives S = 24.

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