Printable

2001 AMC 10 Problem 22

2001 AMC 10 Problem 22AMC 10 Step by Step

Name: Date: September 13, 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 lines that share an unknown: left column versus main diagonal gives x = 22, top row versus left column gives w = 19 and the magic sum 66.

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 $\boxed{\textbf{(D)}\ 46}$ .

Why this works

A magic square gives eight linear equations in six unknowns (five letters and the sum), far more than needed. The efficient move is to subtract two lines that share a letter so that letter and $S$ both disappear, producing a numeric value at once. Two such pairings plus one line to fix $S$ finish the problem.

Alternative approach

In any $3 \times 3$ magic square the center equals $S/3$ (add the middle row, middle column and both diagonals: $4S = 3S + 3x$ ). With $x = 22$ , $S = 66$ , and the middle row and column give $y = 66 - 18 - 22 = 26$ and $z = 66 - 24 - 22 = 20$ directly.

The trap

Writing all eight equations and solving blindly; instead pair lines that share a variable so the unknown cancels immediately.

Common mistakes

Writing all eight equations and solving blindly; instead pair lines that share a variable so the unknown cancels immediately.
Answering $2v = 46$ by luck of the setup but computing $v$ wrongly, or reporting $y$ or $z$ alone.

Techniques

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