2019 AMC 10A Problems/Problem 19


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 1

Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$, which can be simplified to $(x^2+5x+5)^2-1+2019$. Noting that squares are nonnegative, and verifying that $x^2+5x+5=0$ for some real $x$, the answer is $\boxed{\textbf{(B) } 2018}$.

Solution 2

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

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

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

Solution 3 (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:





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

Note: We could also have used the result that minimum/maximum point of a parabola $y = ax^2 + bx + c$ occurs at $x=-\frac{b}{2a}$.

Note 2: This solution is somewhat "lucky", since when we define variables to equal a function, and create another function out of these variables, the domain of such function may vary from the initial one. This is important because the maximum and minimum value of a function is dependent on its domain, e.g:

$f(x)=x^2$ has no maximum value in the the integers, but once restricting the domain to $(-5, 5)$ the maximum value of $f(x)$ is $25$.

Also, observe that if we were to evaluate this by taking the derivative of $(x+1)(x+2)(x+3)(x+4)+2019$, we would get $-5$ as the $x$-value to obtain the minimum $y$-value of this expression. It can be seen that $-5$ is actually an inflection point, instead of a minimum or maximum.

-Note 2 from Benedict T (countmath1)

Solution 4(guess with answer choices)

The expression is negative when an odd number of the factors are negative. This happens when $-2 < x < -1$ or $-4 < x < -3$. Plugging in $x = -\frac32$ or $x = -\frac72$ yields $-\frac{15}{16}$, which is very close to $-1$. Thus the answer is $-1 + 2019 = \boxed{\textbf{(B) }2018}$.

Solution 5 (using the answer choices)

Answer choices $C$, $D$, and $E$ are impossible, since $(x+1)(x+2)(x+3)(x+4)$ can be negative (as seen when e.g. $x = -\frac{3}{2}$). Plug in $x = -\frac{3}{2}$ to see that it becomes $2019 - \frac{15}{16}$, so round this to $\boxed{\textbf{(B) }2018}$.

We can also see that the limit of the function is at least -1 since at the minimum, two of the numbers are less than 1, but two are between 1 and 2.

Solution 6 (no words)

$\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+2019$ $=\left(x+2\right)\left(x+3\right)\left(x+1\right)\left(x+4\right)+2019$ $=\left(x^{2}+5x+6\right)\left(x^{2}+5x+4\right)+2019$ $=\left(\left(x^{2}+5x+5\right)+1\right)\left(\left(x^{2}+5x+5\right)-1\right)+2019$ $=\left(x^{2}+5x+5\right)^{2}-1^{2}+2019$.

$x^{2}+5x+5=0 \Rightarrow x=\frac{-5\pm\sqrt{25-20}}{2}=\frac{-5\pm\sqrt{5}}{2}$.


$0-1^{2}+2019=2018 \Rightarrow \boxed{B}$.

Solution 7(naive solution)

Since we can obviously have the first part of the equation to be negative, let $x$ be -$3.5$. Calculating, we find that it is a little more than 2018, and since we're given the choice of $2018$, we guess that it is $2018$.

Video Solutions

https://www.youtube.com/watch?v=Vf2LkM7ExhY by SpreadTheMathLove

https://www.youtube.com/watch?v=Lis8yKT9WXc (less than 2 minutes)

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS