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2001 AMC 10 Problem 22

2001 AMC 10 Problem 22AMC 10 Step by Step

Name: Date: October 28, 2026

Systems of Equations · difficulty 4 of 5 · about 4 min · position P21-25

In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$ , $w$ , $x$ , $y$ , and $z$ . Find $y + z$ .

A) $43$
B) $44$
C) $45$
D) $46$
E) $47$

Answer

Key insight

Compare two lines that share an unknown number; that unknown cancels. The left column against the top row gives w = 19, and against the diagonal gives x = 22.

Solution

1. Read the square from the picture:
$\begin{array}{ccc} v & 24 & w \\ 18 & x & y \\ 25 & z & 21 \end{array}$
Every row, column and diagonal has the same total.
2. The left column has only one unknown, $v$ : its total is $v + 18 + 25 = v + 43$ . Compare it with other lines through $v$ , so the $v$ cancels.
3. Top row against left column. Top row: $v + 24 + w$ . Left column: $v + 43$ . The totals are equal, so $24 + w = 43$ and $w = 19$ .
4. Diagonal against left column. The diagonal from top left to bottom right is $v + x + 21$ . Setting it equal to $v + 43$ gives $x + 21 = 43$ , so $x = 22$ .
5. Find the common total. The other diagonal is $w + x + 25 = 19 + 22 + 25 = 66$ . So every line adds to $66$ .
6. Bottom row: $25 + z + 21 = 66$ , so $z = 20$ .
7. Middle row: $18 + 22 + y = 66$ , so $y = 26$ .
8. Check with a column: the right column is $w + y + 21 = 19 + 26 + 21 = 66$ . It works.
9. So $y + z = 26 + 20 = 46$ .

The answer is $\boxed{\textbf{(D)}\ 46}$ .

Why this works

A magic square gives many more equations than you need. The quick way is to set two lines equal when they share an unknown: that unknown appears on both sides and cancels, leaving a number. Two such comparisons plus one line to find the common total are enough here.

Alternative approach

In every $3 \times 3$ magic square, the center number is one third of the common total. Once step 4 gives $x = 22$ , the total is $3 \times 22 = 66$ . Then the middle row gives $y = 66 - 18 - 22 = 26$ , and the middle column gives $z = 66 - 24 - 22 = 20$ , so $y + z = 46$ .

The trap

Writing down all eight line equations and trying to solve them all at once, instead of pairing lines that share an unknown so it cancels right away.

Concise solution

The square, row by row, is
$\begin{array}{ccc} v & 24 & w \\ 18 & x & y \\ 25 & z & 21 \end{array}$
Let $S$ be the common sum.

The left column and the main diagonal both contain $v$ : $v + 18 + 25 = v + x + 21$ , so $x = 22$ .

The top row and the left column both contain $v$ : $v + 24 + w = v + 43$ , so $w = 19$ . Now the other diagonal gives $S = w + x + 25 = 19 + 22 + 25 = 66$ .

Bottom row: $25 + z + 21 = 66$ , so $z = 20$ . Middle row: $18 + 22 + y = 66$ , so $y = 26$ . (Then $v = 66 - 43 = 23$ , and every line checks.)

Hence $y + z = 26 + 20 = 46$ .

The answer is $\boxed{\textbf{(D)}\ 46}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Intermediate Algebra, Chapter 1: Basic Techniques for Solving Equations (§1.4)

Techniques

Set up the equation/formula and compute; no special trick needed