AMC 10 · 2023 · #14

Grade 11 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

polynomial-rootsvieta-formulasrational-root-theoremfactorssystematic-enumeration work-backwardssystematic-enumerationcasework ↑ Prerequisites: polynomial-factoringfactors

📏 Long solution 💡 3 insights

Problem

A monic cubic has constant term six; only its two middle coefficients are free and both must be integers. Count how many choices of that pair give the cubic three distinct integer roots.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Make a Systematic List

This is a 'how many' question with no bound on a or b to prune with, so the answer is simply the length of a list — which makes Tool #2 (Make a Systematic List) the whole job, provided the list is complete and duplicate-free. Two preparations make such a list possible. Tool #4 (Introduce a Variable) names the roots p,q,r, and Tool #11 (Work Backwards) reverses the direction of the problem: rather than picking coefficients and hunting for roots, pick the roots and let the expanded product hand back a and b. Then Tool #3 (Eliminate Possibilities) shrinks the pool of possible roots to the eight divisors of 6 before any listing begins, and Tool #9 (Solve an Easier Related Problem) makes the listing itself easy by first solving the sign-free version — write 6 as a product of three positive integers — and attaching minus signs afterwards.

1STEP 1

Narrow the possible roots

A root must divide the constant.

6=-r(r²+ar+b) → r ∣ 6 → r∈{± 1,± 2,± 3,± 6}

2STEP 2

Rebuild the cubic from its roots

Rebuild the cubic from the roots.

x³+ax²+bx+6=(x-p)(x-q)(x-r)

3STEP 3

Match the coefficients

Matching coefficients fixes the product.

(x-p)(x-q)(x-r)=x³-(p+q+r)x²+(pq+qr+rp)x-pqr → a=-(p+q+r), b=pq+qr+rp, pqr=-6

4STEP 4

Fix the sizes first

Strip the signs and fix the sizes.

|p| |q| |r|=6 → {|p|,|q|,|r|}={1,1,6} or {1,2,3}

5STEP 5

Put the signs back

Attach signs so the product matches.

{-1,2,3}, {1,-2,3}, {1,2,-3}, {-1,-2,-3}, {-1,1,6}

6STEP 6

Read the coefficients off each set

Each root set gives one coefficient pair.

(-6,-1), (-4,1), (-2,-5), (0,-7), (6,11)

7STEP 7

Check for collisions and count

None collide, so there are 5.

5 root sets ⇔ 5 pairs (a,b) → (A)

Answer

An independent sweep confirms the count. Every root divides 6, so all three come from the eight numbers ± 1,± 2,± 3,± 6; checking all C(8, 3)=56 three-element subsets, exactly five have product -6, and they are the five already found. Spot-checking one: the roots -1,1,6 give (x+1)(x-1)(x-6)=x³-6x²-x+6, whose constant term is indeed 6, with a=-6 and b=-1 as computed. Five out of 56 is also a believable share, since a random triple of divisors rarely multiplies to exactly -6. The wrong choices are the shapes of specific slips. Ignore the word 'distinct' and two more triples qualify, {1,1,-6} and {-1,-1,-6}, which give the extra pairs (4,-11) and (8,13) and a total of 7 — choice (D); notice only one of those two and the count is 6 — choice (B). Insist instead that the three sizes |p|,|q|,|r| be different from each other, something the problem never asks, and the shape 1 · 1 · 6 vanishes, leaving only 4 — choice (E). Choice (C) 8 is the number of candidate roots, not the number of pairs.

💡Key takeaway

When the coefficients are allowed to be anything, stop searching them — build the cubic out of its roots instead, and let the one fixed coefficient tell you which roots are even possible.

Only divisors of 6 can be roots
Name the roots and rebuild the cubic
Match coefficients, flip the search
Strip the signs first
Put the signs back
Read off (a,b) from each set
Check for collisions and count