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AMC 8 · 1999 · #11

Grade 6 arithmetic
Archetype Pair-Sum Invariant
pattern-recognitionset-partitionsystematic-enumerationsequences-arithmetic identify-subproblemspattern-recognition ↑ Prerequisites: multi-digit-arithmetic
📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is

Pick an answer.

(A)
20
(B)
21
(C)
22
(D)
24
(E)
30
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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Five numbers to place: $1, 4, 7, 10, 13$ (each used exactly once); Shape is a plus sign — one horizontal row of three squares and one vertical column of three squares; The center square belongs to BOTH the row and the column; Row sum $=$ column sum (call this common value $S$); Answer choices: (A) $20$, (B) $21$, (C) $22$, (D) $24$, (E) $30$

Unknowns: The largest possible value of $S$

Understand

Restated: 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.

Givens: Five numbers to place: $1, 4, 7, 10, 13$ (each used exactly once); Shape is a plus sign — one horizontal row of three squares and one vertical column of three squares; The center square belongs to BOTH the row and the column; Row sum $=$ column sum (call this common value $S$); Answer choices: (A) $20$, (B) $21$, (C) $22$, (D) $24$, (E) $30$

Plan

Primary tool: #11 Find an Invariant

Secondary: #1 Draw a Diagram

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 $2S = 35 + \text{center}$ — 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.

Execute — Answer: D

#1 Draw a Diagram 4.OA.A.3 Step 1
  • Sketch the plus sign and mark which square is shared.
  • The horizontal row has three squares (left, center, right).
  • The vertical column has three squares (top, center, bottom).
  • The center square sits inside both.
$$\text{shared square} = \text{center}$$

💡 Drawing the cross shape lets you see at a glance that one square does double duty. Without the picture, it's easy to forget the center counts twice.

#11 Find an Invariant 6.EE.A.2 Step 2
  • Add the row sum and the column sum.
  • Each of the four outside numbers (top, bottom, left, right) is counted once, and the center number is counted twice.
  • Total of all five numbers is $1 + 4 + 7 + 10 + 13 = 35$.
$$S + S = (1 + 4 + 7 + 10 + 13) + \text{center} = 35 + \text{center}$$

💡 Counting each square's contribution to (row sum) + (column sum) shows the center is the only square hit twice. The fixed total $35$ is the invariant — it doesn't depend on which number goes where.

#11 Find an Invariant 6.EE.B.7 Step 3
  • Solve the invariant equation for $S$.
  • Divide both sides by $2$.
$$2S = 35 + \text{center} \;\Rightarrow\; S = \dfrac{35 + \text{center}}{2}$$

💡 Now $S$ depends only on the center number. The four arm squares can be arranged in many ways, but the common sum is locked once the center is chosen.

#11 Find an Invariant 6.EE.B.7 Step 4
  • To make $S$ as large as possible, make the center as large as possible.
  • The largest available number is $13$, so put $13$ in the center.
  • Then $S = (35 + 13) / 2 = 48 / 2 = 24$.
  • Also check that $S$ is a whole number — it is, because $35 + 13 = 48$ is even.
$$\text{center} = 13 \;\Rightarrow\; S = \dfrac{35 + 13}{2} = \dfrac{48}{2} = 24$$

💡 Bigger center means bigger $S$, so go for the biggest number. Need to be sure $35 + \text{center}$ is even — it is for $13$, and that gives a clean integer answer.

#1 Draw a Diagram 4.OA.A.3 Step 5
  • Verify by actually placing the remaining numbers $\{1, 4, 7, 10\}$ so that each arm pair sums to $S - 13 = 24 - 13 = 11$.
  • The pairs $(1, 10)$ and $(4, 7)$ each add to $11$, so put $1$ and $10$ on the row arms and $4$ and $7$ on the column arms (or vice versa).
  • Both arms then total $13 + 11 = 24$.
$$\text{row: } 1 + 13 + 10 = 24,\ \text{column: } 4 + 13 + 7 = 24 \;\Rightarrow\; \textbf{(D)}$$

💡 The arrangement exists, so $S = 24$ is actually reachable — not just an upper bound from the equation.

[1] #1 4.OA.A.3 Sketch the plus sign and mark which square is shared. The horizontal row has thr
[2] #11 6.EE.A.2 Add the row sum and the column sum. Each of the four outside numbers (top, botto
[3] #11 6.EE.B.7 Solve the invariant equation for $S$. Divide both sides by $2$.
[4] #11 6.EE.B.7 To make $S$ as large as possible, make the center as large as possible. The larg
[5] #1 4.OA.A.3 Verify by actually placing the remaining numbers $\{1, 4, 7, 10\}$ so that each

Review

Reasonableness: Check the other choices against the invariant $S = (35 + \text{center}) / 2$. Center $= 1 \Rightarrow S = 18$; center $= 7 \Rightarrow S = 21$; center $= 13 \Rightarrow S = 24$. Centers $4$ and $10$ make $35 + \text{center}$ 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 $\tfrac{35 + 13}{2} = 24$, so no center makes it reachable — confirming (D).

Alternative: Tool #6 (Guess and Check) on the center: try center $= 13$ first because it is the largest. Need to split the remaining four numbers $\{1, 4, 7, 10\}$ into two pairs with equal sums. Their total is $22$, so each pair must sum to $11$: $(1, 10)$ and $(4, 7)$. Both row and column then total $13 + 11 = 24$. Since center $13$ works and any larger center is impossible, $24$ is the maximum — same answer (D).

CCSS standards used (min grade 6)

  • 4.OA.A.3 Solve multistep word problems posed with whole numbers (Drawing the plus shape, identifying the shared center, and checking the final placement by adding the row and column.)
  • 6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Writing the invariant equation $2S = 35 + \text{center}$ that expresses the row sum plus the column sum in terms of the center value.)
  • 6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Solving $2S = 35 + \text{center}$ for $S$ and evaluating it at $\text{center} = 13$ to get $S = 24$.)

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

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

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