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2001 AMC 10 Problem 10

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2001 AMC 10 Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Systems of Equations · difficulty 2 of 5 · about 2 min · position P1-10

If $x$ , $y$ , and $z$ are positive with $xy = 24$ , $xz = 48$ , and $yz = 72$ , then $x + y + z$ is

A)
$18$
B)
$19$
C)
$20$
D)
$22$
E)
$24$

Answer

Key insight

Dividing two of the equations cancels a variable: xz/xy = 2 gives z = 2y, and then yz = 2y^2 = 72 pins down y = 6.

Solution

Divide the second equation by the first: $\dfrac{xz}{xy} = \dfrac{48}{24}$ , so $\dfrac{z}{y} = 2$ and $z = 2y$ .

Substitute into the third equation: $y \cdot 2y = 72$ , so $y^2 = 36$ and $y = 6$ (positive). Then $z = 12$ , and from $xy = 24$ we get $x = 4$ .

Therefore $x + y + z = 4 + 6 + 12 = 22$ .

The answer is $\boxed{\textbf{(D)}\ 22}$ .

Why this works

Each equation is a product of two of the three unknowns, so ratios of equations cancel one variable cleanly. This is the multiplicative analogue of subtracting linear equations. The positivity condition matters: it selects $y = 6$ rather than $-6$ .

Alternative approach

Multiply all three: $(xyz)^2 = 24 \cdot 48 \cdot 72 = 82944 = 288^2$ , so $xyz = 288$ . Dividing by each given product: $z = 288/24 = 12$ , $y = 288/48 = 6$ , $x = 288/72 = 4$ . Same sum, $22$ .

The trap

Taking xyz = 24*48*72 instead of its square root, or a sign slip when the problem says all variables are positive.

Common mistakes

Taking xyz = 244872 instead of its square root, or a sign slip when the problem says all variables are positive.
Guessing a factor triple that fits two equations but not the third (for instance $x = 3, y = 8$ satisfies $xy = 24$ but then $xz = 48$ forces $z = 16$ and $yz = 128$ ).

Techniques

Set up the equation/formula and compute; no special trick needed · Exploit symmetry to reduce work or pair up objects