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

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

Revision as of 19:09, 9 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 $\boxed{(B) 2018}$

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 10 Problems and Solutions

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