2019 AMC 10A Problem 19

Question

What is the least possible value of

\begin{aligned} &(x+1)(x+2)(x+3)(x+4)\\ &\quad+2019, \end{aligned}

where

x

is a real number?

2017

2018

2019

2020

2021

Pair factors symmetrically around

-\frac52

Rewrite the quartic using

u=x^2+5x

Multiplying the first two terms and the last terms yields

(x^2 + 5x + 4)(x^2 + 5x + 6).

Note that these two terms differ by

2.

We can try to express this as a difference of squares, which is

(x^2 + 5x + 5)^2 - 1.

Adding

2019

to this gets us

(x^2 + 5x + 5)^2 + 2018.

Squares are non-negative, so as long as we find a way to make the inner expression

0,

we can make the square

0.

The discriminant is

5^2 - 4 \cdot 5 = 5,

which is positive meaning that there is a value that makes the square

0.

This means that the minimum value would be

0^2 + 2018 = 2018.

Thus, B is the correct answer.

Problem 19 in other years