2019 AMC 10A Problem 19

Below is the professionally curated solution for Problem 19 of the 2019 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 10A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:difference of squaresoptimizationsubstitution

Difficulty rating: 1540

19.

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

20172017

20182018

20192019

20202020

20212021

Solution:

Multiplying the first two terms and the last terms yields (x2+5x+4)(x2+5x+6). (x^2 + 5x + 4)(x^2 + 5x + 6).

Note that these two terms differ by 2.2. We can try to express this as a difference of squares, which is (x2+5x+5)21. (x^2 + 5x + 5)^2 - 1.

Adding 20192019 to this gets us (x2+5x+5)2+2018. (x^2 + 5x + 5)^2 + 2018.

Squares are non-negative, so as long as we find a way to make the inner expression 0,0, we can make the square 0.0.

The discriminant is 5245=5,5^2 - 4 \cdot 5 = 5, which is positive meaning that there is a value that makes the square 0.0.

This means that the minimum value would be 02+2018=2018. 0^2 + 2018 = 2018. Thus, B is the correct answer.

Problem 19 in Other Years