AMC 10 · 2019 · #19

Grade 8 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

polynomial-substitutiondifference-of-squaresperfect-squarespolynomial-factoring identify-subproblemspattern-recognition ↑ Prerequisites: polynomial-substitutiondifference-of-squares

📏 Short solution 💡 2 insights

Problem

For every real number x, compute (x+1)(x+2)(x+3)(x+4) + 2019. Find the smallest value this expression can take.

Pick an answer.

(A)

2017

(B)

2018

(C)

2019

(D)

2020

(E)

2021

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Organize Information in More Ways

Tool #15 (Reorganize): the four factors look messy, but pairing (x+1)(x+4) with (x+2)(x+3) makes both products equal to x² + 5x + (a constant) — the same hidden quantity in each. Tool #7 (Subproblems): once the products share a structure, substitute u = x² + 5x to reduce a quartic-looking expression to a quadratic. Tool #13 (Algebra): finish with a perfect square.

1STEP 1

Pair (x+1)(x+4) = x² + 5x + 4 and (x+2)(x+3) = x² + 5x + 6 — both hide the same chunk x² + 5x.

(x+1)(x+4) = x² + 5x + 4, (x+2)(x+3) = x² + 5x + 6

2STEP 2

Set u = x² + 5x + 5 (midpoint of +4 and +6): the product becomes (u-1)(u+1) = u² - 1.

(x+1)(x+2)(x+3)(x+4) = (u-1)(u+1) = u² - 1

3STEP 3

Add 2019: u² - 1 + 2019 = u² + 2018. Since u² ≥ 0, the minimum is 2018 at u = 0.

u² + 2018 ≥ 2018, equality at u = 0

4STEP 4

Check u = 0 is reachable: x² + 5x + 5 = 0 has real roots x = $\frac{-5 ± √5}{2}$ , so 2018 is truly attained. Answer (B).

x =

\frac{-5 ± √(5)}{2}

→ min = 2018 → (B)

Answer

2018

Plug in x = -2.5 (the symmetry axis): factors become -1.5 · -0.5 · 0.5 · 1.5 = (1.5 · 0.5)² = 0.5625, so expression = 0.5625 + 2019 = 2019.5625 — above the minimum 2018, as expected. At x =

\frac{-5 + √(5)}{2}

≈ -1.382: factors are roughly -0.382, 0.618, 1.618, 2.618; product ≈ -0.382 · 0.618 · 1.618 · 2.618 ≈ -1.000; add 2019 → 2018. Matches. (A) 2017 is the trap of forgetting the -1 went into u² - 1 + 2019 = u² + 2018 (not u² + 2017).

💡Key takeaway

This AMC 10 problem only needs Grade 8 square-and-shift you already know — pair (x+1)(x+4) with (x+2)(x+3), substitute u = x² + 5x + 5 to get u² - 1 + 2019 = u² + 2018, and since u² ≥ 0 the minimum is 2018. The answer is (B).

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