AMC 10 · 2025 · #9

Grade 7 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

symmetry-argumentcaseworkcomplementary-counting bound-inequality-then-enumerate ↑ Prerequisites: systematic-enumeration

📏 Medium solution 💡 3 insights

Problem

How many ordered triples of integers $(x, y, z)$ satisfy the following system of inequalities?
\begin{align*} -x-y-z&\le -2\ -x+y+z&\le 2\ x-y+z&\le 2\ x+y-z&\le 2 \end{align*}

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2025 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: Count how many ordered triples of integers $(x, y, z)$ make all four of these inequalities true at once: $-x-y-z\le -2$, $-x+y+z\le 2$, $x-y+z\le 2$, and $x+y-z\le 2$.

Givens: The system of four inequalities $-x-y-z\le -2,\ -x+y+z\le 2,\ x-y+z\le 2,\ x+y-z\le 2$; $x, y, z$ are integers; Answer choices: (A) $4$, (B) $8$, (C) $11$, (D) $15$, (E) $17$

Unknowns: The number of ordered integer triples $(x, y, z)$ that satisfy all four inequalities

Understand

Restated: Count how many ordered triples of integers $(x, y, z)$ make all four of these inequalities true at once: $-x-y-z\le -2$, $-x+y+z\le 2$, $x-y+z\le 2$, and $x+y-z\le 2$.

Givens: The system of four inequalities $-x-y-z\le -2,\ -x+y+z\le 2,\ x-y+z\le 2,\ x+y-z\le 2$; $x, y, z$ are integers; Answer choices: (A) $4$, (B) $8$, (C) $11$, (D) $15$, (E) $17$

Plan

Primary tool: #2 Make a Systematic List

Secondary: #13 Convert to Algebra, #16 Change Focus / Count the Complement

The question is a pure "how many" count, which is the signature of Tool #2 (Make a Systematic List). But before listing, the search space looks infinite, so Tool #13 (Convert to Algebra) does the trapping: combine the inequalities to prove every variable is squeezed into $0 \le x, y, z \le 2$. That shrinks the problem to a $3\times 3\times 3$ box of only $27$ candidates. Then Tool #2 lists the valid value-sets and counts their orderings. Tool #16 (Count the Complement) gives an independent second count in review — instead of listing what works, throw away the few triples that break a rule.

Execute — Answer: C

#13 Convert to Algebra 7.NS.A.1 Step 1

Flip the first inequality

The first inequality has every variable negative, which is awkward.
Multiply both sides by $-1$; the inequality sign flips.
Now the whole system reads with $x+y+z$ on one side.

$$-x-y-z \le -2 \;\Longrightarrow\; x+y+z \ge 2$$

💡 Multiplying an inequality by a negative reverses the arrow, the way weighing the opposite of everything flips which side is heavier.

#13 Convert to Algebra 7.EE.B.4 Step 2

Trap each variable in 0 to 2

Combine inequalities to bound one variable, say $x$.
Subtracting the second from the first isolates $2x$ on the low end; adding the third and fourth isolates $2x$ on the high end.
The same trick works for $y$ and $z$ by symmetry, so every variable is caught between $0$ and $2$.

$$(x{+}y{+}z)-(-x{+}y{+}z)=2x\ge 0,\quad (x{-}y{+}z)+(x{+}y{-}z)=2x\le 4 \;\Rightarrow\; 0\le x\le 2,\ \text{likewise } 0\le y,z\le 2$$

💡 Adding or subtracting two true inequalities makes a new true inequality, and the right pairing cancels two variables so only one is left to bound.

#2 Make a Systematic List 7.SP.C.8 Step 3

Reduce to 27 candidates

Because $x, y, z$ are integers with $0 \le x, y, z \le 2$, each variable can only be $0$, $1$, or $2$.
That is $3$ choices each, so at most $3 \times 3 \times 3 = 27$ triples to check — a finite list we can actually handle.

$$x, y, z \in \{0, 1, 2\} \;\Rightarrow\; 3\times 3\times 3 = 27 \text{ candidate triples}$$

💡 Once a variable is boxed into a short range of whole numbers, an endless problem becomes a small pile you can count.

#2 Make a Systematic List 6.EE.B.5 Step 4

Keep only the value-sets that fit

In the box the four rules become $x+y+z \ge 2$ and each of $y+z-x,\ x+z-y,\ x+y-z \le 2$.
Test value-sets by substitution.
A sum below $2$ dies; any two big values with a small third makes a difference of $3$ or more and dies.
The surviving unordered value-sets are just five.

$$\text{Valid multisets: } \{0,0,2\},\ \{0,1,1\},\ \{1,1,1\},\ \{1,1,2\},\ \{2,2,2\}$$

💡 Checking a candidate is just plugging its numbers in and asking "do all four still read true?"

#2 Make a Systematic List 7.SP.C.8 Step 5

Count every ordering

The question wants ordered triples, so each value-set contributes one triple for every distinct arrangement of its numbers.
A set with three different-looking slots like $\{1,1,2\}$ has $3$ orderings (the $2$ can sit in any of three spots); all-same sets like $\{1,1,1\}$ have just $1$.
Add them up.

$$\underbrace{3}_{\{0,0,2\}}+\underbrace{3}_{\{0,1,1\}}+\underbrace{1}_{\{1,1,1\}}+\underbrace{3}_{\{1,1,2\}}+\underbrace{1}_{\{2,2,2\}} = 11 \;\Rightarrow\; \textbf{(C)}$$

💡 Ordered counting means the same three numbers in a different order is a brand-new answer, so multiply out the arrangements of each set.

[1] #13 7.NS.A.1 The first inequality has every variable negative, which is awkward. Multiply bot

[2] #13 7.EE.B.4 Combine inequalities to bound one variable, say $x$. Subtracting the second from

[3] #2 7.SP.C.8 Because $x, y, z$ are integers with $0 \le x, y, z \le 2$, each variable can onl

[4] #2 6.EE.B.5 In the box the four rules become $x+y+z \ge 2$ and each of $y+z-x,\ x+z-y,\ x+y-

[5] #2 7.SP.C.8 The question wants ordered triples, so each value-set contributes one triple for

Review

Reasonableness: Independent recount by complement (Tool #16): start from all $27$ candidates in the box. Drop the ones with $x+y+z < 2$: that is $(0,0,0)$ plus the three orderings of $(1,0,0)$, so $4$ triples. Drop the ones where a difference exceeds $2$: for the rule $y+z-x\le 2$ the offenders are $(0,1,2),(0,2,1),(0,2,2),(1,2,2)$, which is $4$; by symmetry the other two difference-rules kill $4$ each, and none of these $12$ overlap each other or the first group. So valid $= 27 - 4 - 12 = 11$. Both methods land on $11$, matching (C).

Alternative: The complement count above IS the clean alternative: rather than listing what survives, subtract what fails. It avoids sorting triples into value-sets and instead just removes the $4$ low-sum triples and the $12$ big-difference triples from the $27$-triple box, giving $27-4-12=11$ directly.

CCSS standards used (min grade 7)

7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers (Multiplying the first inequality by $-1$ and flipping its direction to rewrite $-x-y-z\le-2$ as $x+y+z\ge 2$.)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities (Adding and subtracting the given inequalities to bound each variable into $0\le x,y,z\le 2$.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Testing candidate value-sets by substitution to see which ones satisfy all four inequalities.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Enumerating the 27-candidate sample space and counting the ordered arrangements of each valid value-set to total 11.)

⭐ Trap each variable into $0, 1,$ or $2$ first, then the endless-looking problem is just a tiny list of $27$ you can check and count.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Flip the first inequality

Trap each variable in 0 to 2

Reduce to 27 candidates

Keep only the value-sets that fit

Count every ordering

Review

CCSS standards used (min grade 7)

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