Difference between revisions of "2019 AMC 10A Problems/Problem 19"

Revision as of 01:28, 10 February 2019

Problem

What is the least possible value of $(x+1)(x+2)(x+3)(x+4)+2019$ where $x$ is a real number?

$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$

Solution

Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$ which can be simplified as $(x^2+5x+5)^2-1+2019$ . Since squares are nonnegative, the answer is $\boxed{(B) 2018}$

Solution 2

Let $a=x+\tfrac{5}{2}$ . Then $(x+1)(x+2)(x+3)(x+4)$ becomes $(a-\tfrac{3}{2})(a-\tfrac{1}{2})(a+\tfrac{1}{2})(a+\tfrac{3}{2})$

We can use difference of squares to get $(a^2-\tfrac{9}{4})(a^2-\tfrac{1}{4})$ , and expand this to get $a^4-\tfrac{5}{2}a+\tfrac{9}{16}$ .

Refactor this by completing the square to get $(a^2-\tfrac{5}{4})^2-1$ , which has a minimum value of $-1$ . The answer is thus $2019-1=\boxed{2018}$

-WannabeCharmander

Solution 3 (using calculus)

Similar to Solution 1, grouping the first and last terms and the middle terms, we get $(x^2+5x+4)(x^2+5x+6)+2019$ .

Letting $y=x^2+5x$ , we get the expression $(y+4)(y+6)+2019$ . Now, we can find the critical points of $(y+4)(y+6)$ to minimize the function.

$\frac{d}{dx}(y^2+10y+24)=0$

$2y+10=0$

$2y(y+5)=0$

$y=-5,0$

To minimize the result, we use $y=-5$ . Hence, the minimum is $(-5+4)(-5+6)=-1$ , so $-1+2019 = 2018 \rightarrow \boxed{B}$

(inspired by solution by oO8715_alexOo)

Note: The minimum/maximum of a parabola occurs at $x=-b/2a

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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