Difference between revisions of "2019 AMC 10A Problems/Problem 19"
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==Solution== | ==Solution== | ||
| + | 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 18:09, 9 February 2019
Problem
What is the least possible value of
![\[(x+1)(x+2)(x+3)(x+4)+2019\]](http://latex.artofproblemsolving.com/1/2/8/128007ee3b97f3ff741a1e668139edf8c13ba7e1.png)
where 
is a real number?
Solution
Grouping the first and last terms and two middle terms gives 
which can be simplified as 
. Since squares are nonnegative, the answer is 
See Also
| 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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