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2022 AMC 10B Problem 18

2022 AMC 10B Problem 18AMC 10 Step by Step

Name: Date: September 13, 2026

Systems of Equations · difficulty 4 of 5 · about 8 min · position P16-20

Consider systems of three linear equations with unknowns $x$ , $y$ , and $z$ , \begin{align} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ . For example, one such system is $\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}$ with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there? (The equations in a system need not be distinct, and two systems containing the same equations in a different order are considered different.)

A)
$302$
B)
$338$
C)
$340$
D)
$343$
E)
$344$

Answer

Key insight

A nonzero solution exists iff the rows are dependent: a zero row, a repeated row, or three distinct rows with one equal to the sum of the others.

Solution

Think of each equation as its coefficient row, one of the $8$ vectors in $\{0,1\}^3$ ; a system is an ordered triple of rows, $8^3 = 512$ in all. Three homogeneous equations have a nonzero solution exactly when the rows are linearly dependent, i.e. one row is a combination of the other two (each equation is a plane through the origin; three such planes share a line iff their normals are coplanar).

Count the dependent triples in three exclusive cases.

A zero row. Then only two real equations remain, and two equations in three unknowns always have a nonzero solution. Triples with at least one row $000$ : $8^3 - 7^3 = 512 - 343 = 169$ .

No zero row, a repeated row. Again only two distinct equations. Ordered triples of nonzero rows that are not all distinct: $7^3 - 7 \cdot 6 \cdot 5 = 343 - 210 = 133$ .

Three distinct nonzero rows, dependent. Suppose $w$ is a combination of $u$ and $v$ . Comparing coordinates where $u$ and $v$ have entries $(1,0)$ , $(0,1)$ , $(1,1)$ forces the coefficients into $\{0, \pm1\}$ , and since $w \ne 0, u, v$ , one row must be the sum of the other two. The sum of two $0$ - $1$ rows is again a $0$ - $1$ row only if they share no $1$ s, giving the sets
$\{100, 010, 110\},\ \{100, 001, 101\},\ \{010, 001, 011\},\ \{100, 011, 111\},\ \{010, 101, 111\},\ \{001, 110, 111\}.$
Six sets, each in $3! = 6$ orders: $36$ .

Total: $169 + 133 + 36 = 338$ .

The answer is $\boxed{\textbf{(B)}\ 338}$ .

Why this works

"Has a nonzero solution" is a statement about the equations being redundant, not about solving anything. Sorting the redundancy into "an equation is trivial," "an equation repeats," and "an equation is the sum of the others" makes the count mechanical, and the $0$ - $1$ restriction keeps the last case tiny. Order matters and repeats are allowed, so count ordered triples throughout.

Alternative approach

Count the complement, systems with only the trivial solution, i.e. independent rows. First row: any of $7$ nonzero rows. Second: any of the $6$ others (no two distinct nonzero $0$ - $1$ rows are proportional). The plane spanned by the first two contains $4$ of the $8$ rows ( $0$ , both rows, and their sum or difference), except when both are weight- $2$ rows like $110, 101$ , where it contains only $3$ . So the third row has $4$ choices in $36$ cases and $5$ choices in $6$ cases: $36 \cdot 4 + 6 \cdot 5 = 174$ independent systems, and $512 - 174 = 338$ .

The trap

Counting only systems with a zero row or a repeated row (302) and missing the 36 systems whose three distinct rows satisfy u + v = w.

Common mistakes

Counting only systems with a zero row or a repeated row (302) and missing the 36 systems whose three distinct rows satisfy u + v = w.
Treating systems with the same equations in a different order as identical, or forgetting that a row may repeat.

Techniques

Split into exhaustive cases and handle each · Count the complement and subtract from the total