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

Latest revision as of 02:21, 2 September 2024

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:

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

$2y+10=0$

$2y=-5$

$y=-5,0$

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

Note 2: This solution is somewhat "lucky", since when we define variables to equal a function, and create another function out of these variables, the domain of such function may vary from the initial one. This is important because the maximum and minimum value of a function is dependent on its domain, e.g:

$f(x)=x^2$ has no maximum value in the the integers, but once restricting the domain to $(-5, 5)$ the maximum value of $f(x)$ is $25$ .

Also, observe that if we were to evaluate this by taking the derivative of $(x+1)(x+2)(x+3)(x+4)+2019$ , we would get $-5$ as the $x$ -value to obtain the minimum $y$ -value of this expression. It can be seen that $-5$ is actually an inflection point, instead of a minimum or maximum.

-Note 2 from Benedict T (countmath1)

Solution 4(guess with answer choices)

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 $2$ .

Solution 6 (also calculus but more convoluted)

We can ignore the $2019$ and consider it later, as it is a constant. By difference of squares, we can group this into $\left((x+2.5)^2-0.5^2\right)\left((x+2.5)^2-1.5^2\right)$ . We pull a factor of $4$ into each term to avoid dealing with decimals:

$\dfrac{\left((2x+5)^2-1\right)\left((2x+5)^2-9\right)}{16}.$

Now, we let $a=2x+5$ . Our expression becomes:

$\dfrac{(a-1)(a-9)}{16}=\dfrac{a^2-10a+9}{16}.$

Taking the derivative, we get $\dfrac{2a-10}{16}=\dfrac{a-5}8.$ This is equal to $0$ when $a=5$ , and plugging in $a=5$ , we get the expression is equal to $-1$ and therefore our answer is $2019-1=\boxed{\text{(B)}~2018}.$

~Technodoggo

Video Solutions

https://www.youtube.com/watch?v=Vf2LkM7ExhY by SpreadTheMathLove

https://www.youtube.com/watch?v=Lis8yKT9WXc (less than 2 minutes)

https://youtu.be/NRa3VnjNVbw - Education, the Study of Everything
https://www.youtube.com/watch?v=Mfa7j2BoNjI
https://youtu.be/tIzJtgJbHGc - savannahsolver
https://youtu.be/3dfbWzOfJAI?t=3319 - pi_is_3.14
https://youtu.be/GmUWIXXf_uk?t=1134 ~ pi_is_3.14

@@ Line 6: / Line 6: @@
 <math>\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021</math>
-==Solution==
+==Solution 1==
+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 to <math>(x^2+5x+5)^2-1+2019</math>. Noting that squares are nonnegative, and verifying that <math>x^2+5x+5=0</math> for some real <math>x</math>, the answer is <math>\boxed{\textbf{(B) } 2018}</math>.
+==Solution 2==
+Let <math>a=x+\tfrac{5}{2}</math>. Then the expression <math>(x+1)(x+2)(x+3)(x+4)</math> becomes <math>\left(a-\tfrac{3}{2}\right)\left(a-\tfrac{1}{2}\right)\left(a+\tfrac{1}{2}\right)\left(a+\tfrac{3}{2}\right)</math>.
+We can now use the difference of two squares to get <math>\left(a^2-\tfrac{9}{4}\right)\left(a^2-\tfrac{1}{4}\right)</math>, and expand this to get <math>a^4-\tfrac{5}{2}a^2+\tfrac{9}{16}</math>.
+Refactor this by completing the square to get <math>\left(a^2-\tfrac{5}{4}\right)^2-1</math>, which has a minimum value of <math>-1</math>. The answer is thus <math>2019-1=\boxed{\textbf{(B) }2018}</math>.
+==Solution 3 (calculus)==
+Similar to Solution 1, grouping the first and last terms and the middle terms, we get <math>(x^2+5x+4)(x^2+5x+6)+2019</math>.
+Letting <math>y=x^2+5x</math>, we get the expression <math>(y+4)(y+6)+2019</math>. Now, we can find the critical points of <math>(y+4)(y+6)</math> to minimize the function:
+<math>\frac{d}{dy}(y^2+10y+24)=0</math>
+<math>2y+10=0</math>
+<math>2y=-5</math>
+<math>y=-5,0</math>
+To minimize the result, we use <math>y=-5</math>. Hence, the minimum is <math>(-5+4)(-5+6)=-1</math>, so <math>-1+2019 = \boxed{\textbf{(B) }2018}</math>.
+''Note'': We could also have used the result that minimum/maximum point of a parabola <math>y = ax^2 + bx + c</math> occurs at <math>x=-\frac{b}{2a}</math>.
+''Note 2:'' This solution is somewhat "lucky", since when we define variables to equal a function, and create another function out of these variables, the domain of such function may vary from the initial one. This is important because the maximum and minimum value of a function is dependent on its domain, e.g:
+<math>f(x)=x^2</math> has no maximum value in the the integers, but once restricting the domain to <math>(-5, 5)</math> the maximum value of <math>f(x)</math> is <math>25</math>.
+Also, observe that if we were to evaluate this by taking the derivative of <math>(x+1)(x+2)(x+3)(x+4)+2019</math>, we would get <math>-5</math> as the <math>x</math>-value to obtain the minimum <math>y</math>-value of this expression. It can be seen that <math>-5</math> is actually an inflection point, instead of a minimum or maximum.
+-Note 2 from Benedict T (countmath1)
+==Solution 4(guess with answer choices)==
+The expression is negative when an odd number of the factors are negative. This happens when <math>-2 < x < -1</math> or <math>-4 < x < -3</math>. Plugging in <math>x = -\frac32</math> or <math>x = -\frac72</math> yields <math>-\frac{15}{16}</math>, which is very close to <math>-1</math>. Thus the answer is <math>-1 + 2019 = \boxed{\textbf{(B) }2018}</math>.
+==Solution 5 (using the answer choices) ==
+Answer choices <math>C</math>, <math>D</math>, and <math>E</math> are impossible, since <math>(x+1)(x+2)(x+3)(x+4)</math> can be negative (as seen when e.g. <math>x = -\frac{3}{2}</math>). Plug in <math>x = -\frac{3}{2}</math> to see that it becomes <math>2019 - \frac{15}{16}</math>, so round this to <math>\boxed{\textbf{(B) }2018}</math>.
+We can also see that the limit of the function is at least <math>-1</math> since at the minimum, two of the numbers are less than <math>1</math>, but two are between <math>1</math> and <math>2</math>.
+==Solution 6 (also calculus but more convoluted)==
+We can ignore the <math>2019</math> and consider it later, as it is a constant. By difference of squares, we can group this into <math>\left((x+2.5)^2-0.5^2\right)\left((x+2.5)^2-1.5^2\right)</math>. We pull a factor of <math>4</math> into each term to avoid dealing with decimals:
+<cmath>\dfrac{\left((2x+5)^2-1\right)\left((2x+5)^2-9\right)}{16}.</cmath>
+Now, we let <math>a=2x+5</math>. Our expression becomes:
+<cmath>\dfrac{(a-1)(a-9)}{16}=\dfrac{a^2-10a+9}{16}.</cmath>
+Taking the derivative, we get <math>\dfrac{2a-10}{16}=\dfrac{a-5}8.</math> This is equal to <math>0</math> when <math>a=5</math>, and plugging in <math>a=5</math>, we get the expression is equal to <math>-1</math> and therefore our answer is <math>2019-1=\boxed{\text{(B)}~2018}.</math>
+~Technodoggo
+==Video Solutions==
+https://www.youtube.com/watch?v=Vf2LkM7ExhY by SpreadTheMathLove
+https://www.youtube.com/watch?v=Lis8yKT9WXc (less than 2 minutes)
+*https://youtu.be/NRa3VnjNVbw - Education, the Study of Everything
+*https://www.youtube.com/watch?v=Mfa7j2BoNjI
+*https://youtu.be/tIzJtgJbHGc - savannahsolver
+*https://youtu.be/3dfbWzOfJAI?t=3319 - pi_is_3.14
+*https://youtu.be/GmUWIXXf_uk?t=1134 ~ pi_is_3.14
 ==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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