AMC 10 · 2019 · #19
Grade 8 arithmeticProblem
What is the least possible value of
(x+1)(x+2)(x+3)(x+4)+2019where x is a real number?
Pick an answer.
AMC 10 2019 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.
Toolkit + CCSS Solution
Understand
Restated: For every real number $x$, compute $(x+1)(x+2)(x+3)(x+4) + 2019$. Find the smallest value this expression can take.
Givens: Expression: $(x+1)(x+2)(x+3)(x+4) + 2019$; $x$ ranges over all real numbers; Choices: (A) $2017$, (B) $2018$, (C) $2019$, (D) $2020$, (E) $2021$
Unknowns: The minimum value of the expression over real $x$
Understand
Restated: For every real number $x$, compute $(x+1)(x+2)(x+3)(x+4) + 2019$. Find the smallest value this expression can take.
Givens: Expression: $(x+1)(x+2)(x+3)(x+4) + 2019$; $x$ ranges over all real numbers; Choices: (A) $2017$, (B) $2018$, (C) $2019$, (D) $2020$, (E) $2021$
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #7 Identify Subproblems, #13 Convert to Algebra
Tool #15 (Reorganize): the four factors look messy, but pairing $(x+1)(x+4)$ with $(x+2)(x+3)$ makes both products equal to $x^2 + 5x + (\text{a constant})$ — the same hidden quantity in each. Tool #7 (Subproblems): once the products share a structure, substitute $u = x^2 + 5x$ to reduce a quartic-looking expression to a quadratic. Tool #13 (Algebra): finish with a perfect square.
Execute — Answer: B
6.EE.A.3 Step 1 - Pair the factors cleverly: $(x+1)(x+4) = x^2 + 5x + 4$ and $(x+2)(x+3) = x^2 + 5x + 6$.
- Both contain the same chunk $x^2 + 5x$ — the trick is to make that chunk visible.
💡 Pair outside-with-inside — both products share $x^2 + 5x$.
7.EE.A.1 Step 2 - Let $u = x^2 + 5x + 5$ (the midpoint between $+4$ and $+6$).
- Then $(x+1)(x+4) = u - 1$ and $(x+2)(x+3) = u + 1$.
- The original product becomes a difference of squares: $(u-1)(u+1) = u^2 - 1$.
💡 Symmetric labels around a midpoint $\Rightarrow$ classic $(u-1)(u+1) = u^2 - 1$.
8.EE.A.2 Step 3 - Add the $+ 2019$ to get $u^2 - 1 + 2019 = u^2 + 2018$.
- Since $u^2 \geq 0$ for every real $u$, the minimum of $u^2 + 2018$ is $2018$, achieved when $u = 0$.
💡 A square is $\geq 0$ — so add the constant to read off the floor.
8.EE.A.2 Step 4 - Confirm $u = 0$ is reachable.
- $u = x^2 + 5x + 5 = 0$ gives $x = \frac{-5 \pm \sqrt{25 - 20}}{2} = \frac{-5 \pm \sqrt{5}}{2}$ — real solutions.
- So the minimum $2018$ is actually attained.
- The answer is (B).
💡 The quadratic $u(x) = 0$ has real roots, so the bound $2018$ is achieved.
6.EE.A.3 Pair the factors cleverly: $(x+1)(x+4) = x^2 + 5x + 4$ and $(x+2)(x+3) = x^2 + 5 7.EE.A.1 Let $u = x^2 + 5x + 5$ (the midpoint between $+4$ and $+6$). Then $(x+1)(x+4) = 8.EE.A.2 Add the $+ 2019$ to get $u^2 - 1 + 2019 = u^2 + 2018$. Since $u^2 \geq 0$ for ev 8.EE.A.2 Confirm $u = 0$ is reachable. $u = x^2 + 5x + 5 = 0$ gives $x = \frac{-5 \pm \sq Review
Reasonableness: Plug in $x = -2.5$ (the symmetry axis): factors become $-1.5 \cdot -0.5 \cdot 0.5 \cdot 1.5 = (1.5 \cdot 0.5)^2 = 0.5625$, so expression $= 0.5625 + 2019 = 2019.5625$ — above the minimum $2018$, as expected. At $x = \frac{-5 + \sqrt{5}}{2} \approx -1.382$: factors are roughly $-0.382, 0.618, 1.618, 2.618$; product $\approx -0.382 \cdot 0.618 \cdot 1.618 \cdot 2.618 \approx -1.000$; add $2019 \to 2018$. Matches. (A) $2017$ is the trap of forgetting the $-1$ went into $u^2 - 1 + 2019 = u^2 + 2018$ (not $u^2 + 2017$).
Alternative: Tool #5 (Pattern): expand $(x+1)(x+2)(x+3)(x+4)$ as a quartic in $x$ and take its derivative — find critical points. By symmetry around $x = -2.5$, the minimum sits at $x = -2.5 \pm \sqrt{5}/2$, where the value equals $-1 + 2019 = 2018$. Same answer, more computation.
CCSS standards used (min grade 8)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(x+1)(x+4)$ and $(x+2)(x+3)$ to $x^2 + 5x + 4$ and $x^2 + 5x + 6$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Substituting $u = x^2 + 5x + 5$ and using $(u-1)(u+1) = u^2 - 1$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Showing $u = 0$ has a real solution and reading off the minimum value $2018$.)
⭐ 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 $u = x^2 + 5x + 5$ to get $u^2 - 1 + 2019 = u^2 + 2018$, and since $u^2 \geq 0$ the minimum is $2018$. The answer is $\textbf{(B)}$.
⭐ 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 $u = x^2 + 5x + 5$ to get $u^2 - 1 + 2019 = u^2 + 2018$, and since $u^2 \geq 0$ the minimum is $2018$. The answer is $\textbf{(B)}$.
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