AMC 10 · 2022 · #7

Grade 7 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

polynomial-factoringpolynomial-rootsvieta-formulasfactorssystematic-enumeration systematic-enumerationpattern-recognitionidentify-subproblems ↑ Prerequisites: polynomial-factoringfactors

📏 Medium solution 💡 2 insights

Problem

For how many integer values of the constant k does the quadratic x² + kx + 36 have two distinct integer roots?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The roots have to multiply to 36, so we are really being asked to list every integer factor pair of 36 in which the two factors are different. Tool #2 (Systematic List) is built for 'find every pair' with no duplicates. Tool #7 (Subproblems) splits the work into 'list positive factor pairs', 'list negative factor pairs', and 'count distinct k'. Tool #9 (Easier Problem) is the bridge: instead of solving for k first, solve the simpler factor-pair question and read k off at the end. We avoid the quadratic formula — factor-pair counting is the natural elementary path.

1STEP 1

Write x² + kx + 36 = (x - r)(x - s) and expand: matching coefficients gives rs = 36 and r + s = -k.

x² + kx + 36 = (x-r)(x-s) → rs = 36, r + s = -k

2STEP 2

List the positive factor pairs of 36, smaller factor first: (1, 36), (2, 18), (3, 12), (4, 9); drop (6, 6) since the roots must differ.

positive pairs with r ＜ s: (1, 36), (2, 18), (3, 12), (4, 9)

3STEP 3

Flip both signs (negative × negative = positive): (-1, -36), (-2, -18), (-3, -12), (-4, -9); drop (-6, -6) too.

negative pairs: (-1, -36), (-2, -18), (-3, -12), (-4, -9)

4STEP 4

Mixed-sign pairs can't work — a positive times a negative is negative, never +36 — so no other pairs exist.

(+) × (-) = (-) ≠ 36

5STEP 5

Each pair gives k = -(r + s): positive pairs → negative k, negative pairs → positive k, so k ∈ {-37, -20, -15, -13, 13, 15, 20, 37}.

k ∈ {-37, -20, -15, -13, 37, 20, 15, 13}

6STEP 6

All eight k are distinct — positives and negatives never coincide — so there are 8 values. Answer (B).

|{-37, -20, -15, -13, 13, 15, 20, 37}| = 8 → (B)

Answer

Spot-verify the largest k: if k = -37 the polynomial is x² - 37x + 36 = (x - 1)(x - 36) with roots 1 and 36 ✓. Spot-verify a positive k: if k = 13 then x² + 13x + 36 = (x + 4)(x + 9) with roots -4 and -9 ✓. The structural reason the count is 8 and not 9: the factor pair (6, 6) — which would give k = -12 — is forbidden because the roots must be distinct. The structural reason it is not 14 or 16: pairs like (1, 36) and (36, 1) are the same pair (and give the same k), so unordered pair counting is correct.

💡Key takeaway

This AMC 10 problem only needs Grade 7 expanding (x-r)(x-s) you already know — the constant 36 has to come from r · s, so list the four positive factor pairs of 36 with r ≠ s, double them for negatives, and read off 8 values of k.

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