AMC 10 · 2020 · #7

Grade 6 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

sequences-arithmeticmean-median-mode-rangesystematic-enumeration identify-subproblemseasier-related-problem ↑ Prerequisites: sequences-arithmetic

📏 Short solution 💡 2 insights

Problem

The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Place the $25$ integers from $-10$ through $14$ in a $5 \times 5$ square so every row, every column, and both main diagonals add to the same number. Find that common sum.

Givens: The $25$ numbers placed are exactly $-10, -9, -8, \ldots, 12, 13, 14$ (each used once); Grid is $5$ rows by $5$ columns; Every row sum = every column sum = each main-diagonal sum = some constant $C$; Choices: (A) $2$, (B) $5$, (C) $10$, (D) $25$, (E) $50$

Unknowns: The common sum $C$

Understand

Restated: Place the $25$ integers from $-10$ through $14$ in a $5 \times 5$ square so every row, every column, and both main diagonals add to the same number. Find that common sum.

Plan

Primary tool: #7 Identify Subproblems

Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities

Two clean subproblems hide inside this magic-square question. Sub-A: find the grand total of all $25$ numbers. Sub-B: each row sums to the same $C$, and $5$ rows partition the grid, so grand total $= 5C$. Tool #7 surfaces those two pieces. Tool #9 (easier related) lets us pair $-10$ with $10$, $-9$ with $9$, etc., to compute the total in one mental step instead of summing $25$ values directly. Tool #3 pins the answer down to one of the five choices.

Execute — Answer: C

#9 Solve an Easier Related Problem 6.NS.C.5 Step 1

Compute the grand total of the $25$ numbers by pairing.
The block $-10, -9, \ldots, 9, 10$ has $21$ numbers and adds to $0$ (each positive cancels its negative).
What remains is $11 + 12 + 13 + 14$.

$$(-10) + (-9) + \cdots + 9 + 10 = 0$$

💡 Symmetric pairs around $0$ cancel — fastest way to sum a balanced range.

#9 Solve an Easier Related Problem 4.NBT.B.4 Step 2

Add the leftover four positives.

$$11 + 12 + 13 + 14 = 50$$

💡 Pair $11+14 = 25$ and $12+13 = 25$, then $25+25 = 50$.

#7 Identify Subproblems 3.OA.A.3 Step 3

So the grand total of the $25$ numbers is $50$.
Call the common row sum $C$.
The $5$ rows partition all $25$ cells, so adding the row sums gives the grand total.

$$\text{Grand total} = 5C = 50$$

💡 Five equal slices that fill the whole grid must each be one-fifth of the total.

#7 Identify Subproblems 3.OA.A.3 Step 4

Solve for $C$.

$$C = \dfrac{50}{5} = 10$$

💡 Divide both sides by $5$.

#3 Eliminate Possibilities 4.NBT.A.2 Step 5

$C = 10$ matches choice (C).
The column-sum and diagonal-sum constraints are extra conditions that the arrangement must also meet, but they don't change the value of the common sum — it is forced by the row total alone.

$$C = 10\;\Rightarrow\;\textbf{(C)}$$

💡 Read the matching choice; extra symmetry conditions don't change $C$.

[1] #9 6.NS.C.5 Compute the grand total of the $25$ numbers by pairing. The block $-10, -9, \ldo

[2] #9 4.NBT.B.4 Add the leftover four positives.

[3] #7 3.OA.A.3 So the grand total of the $25$ numbers is $50$. Call the common row sum $C$. The

[4] #7 3.OA.A.3 Solve for $C$.

[5] #3 4.NBT.A.2 $C = 10$ matches choice (C). The column-sum and diagonal-sum constraints are ext

Review

Reasonableness: Cross-check by computing the average. The mean of the $25$ numbers is $\dfrac{50}{25} = 2$, and each row holds $5$ numbers, so a typical row sums to $5 \cdot 2 = 10$. That matches. The numerical center of $-10\ldots14$ is $2$ (since $\dfrac{-10 + 14}{2} = 2$), so $5 \cdot 2 = 10$ is exactly what a balanced row would carry.

Alternative: Tool #9 (Easier Related Problem) on a smaller magic square: the classic $3 \times 3$ magic square using $1\ldots9$ has grand total $45$ and common sum $45/3 = 15$. The pattern 'common sum $=$ grand total $\div$ side length' transfers directly. Apply it: $50 / 5 = 10$. Same answer, derived from a familiar smaller case.

CCSS standards used (min grade 6)

6.NS.C.5 Understand that positive and negative numbers describe quantities (Pairing each negative number with its positive partner to cancel and reduce the sum.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Adding $11 + 12 + 13 + 14 = 50$ to finish the grand total.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Translating 'five equal row sums fill the grid' into $5C = 50$ and dividing to get $C = 10$.)
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Matching the computed value $10$ to choice (C).)

⭐ This AMC 10 problem only needs Grade 6 'negatives cancel positives' plus dividing into equal groups — the $25$ numbers add to $50$, the $5$ row sums all equal the same number $C$, so $5C = 50$ gives $C = 10$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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