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

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<math>\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021</math>
 
<math>\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021</math>
 
== Video Solution==
 
https://youtu.be/3dfbWzOfJAI?t=3319
 
 
~ pi_is_3.14
 
  
 
==Solution 1==
 
==Solution 1==
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==Video Solution 1==
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==Video Solutions==
https://youtu.be/NRa3VnjNVbw
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*https://youtu.be/NRa3VnjNVbw - Education, the Study of Everything
 
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*https://www.youtube.com/watch?v=Mfa7j2BoNjI
Education, the Study of Everything
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*https://youtu.be/tIzJtgJbHGc - savannahsolver
 
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*https://youtu.be/3dfbWzOfJAI?t=3319 - pi_is_3.14
==Video Solution 2==
 
For those who want a video solution: https://www.youtube.com/watch?v=Mfa7j2BoNjI
 
 
 
https://youtu.be/tIzJtgJbHGc
 
 
 
~savannahsolver
 
  
 
==See Also==
 
==See Also==

Revision as of 14:19, 19 January 2021

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 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:

$\frac{d}{dx}(y^2+10y+24)=0$

$2y+10=0$

$2y(y+5)=0$

$y=-5,0$

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 $x=-\frac{b}{2a}$.

Solution 4

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.


Video Solutions

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

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