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

Pick an answer.

(A)
4
(B)
8
(C)
11
(D)
15
(E)
17

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

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 ≤ x, y, z ≤ 2. That shrinks the problem to a 3 × 3 × 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.

1STEP 1

Flip the first inequality

Every variable in the first inequality is negative, so multiply both sides by -1; the sign flips and it reads x+y+z ≥ 2.

-x-y-z ≤ -2 ⟹ x+y+z ≥ 2
2STEP 2

Trap each variable in 0 to 2

Subtracting the second from the first leaves 2x ≥ 0; adding the third and fourth leaves 2x ≤ 4 — by symmetry 0 ≤ x, y, z ≤ 2.

(x+y+z)-(-x+y+z)=2x ≥ 0, (x-y+z)+(x+y-z)=2x ≤ 4 → 0 ≤ x ≤ 2, likewise 0 ≤ y,z ≤ 2
3STEP 3

Reduce to 27 candidates

Boxed between 0 and 2 and forced to be whole numbers, each variable is only 0, 1, or 2 — just 27 triples left to check.

x, y, z ∈ {0, 1, 2} → 3 × 3 × 3 = 27 candidate triples
4STEP 4

Keep only the value-sets that fit

Substitute each candidate: a sum below 2 fails, and two big values beside a small one fails — only five value-sets survive.

Valid multisets: {0,0,2}, {0,1,1}, {1,1,1}, {1,1,2}, {2,2,2}
5STEP 5

Count every ordering

Order matters, so add each set's arrangements — three each for the two-alike sets, one each for the all-alike sets: 3+3+1+3+1 = 11.

3_{{0,0,2}}+3_{{0,1,1}}+1_{{1,1,1}}+3_{{1,1,2}}+1_{{2,2,2}} = 11 → (C)
Answer
11
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 ≤ 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).
💡Key takeaway

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.

  • 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

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