Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AMC 10A Problems/Problem 21"

(→‎Solution 2)
m (→‎Solution 2)
 
(19 intermediate revisions by 9 users not shown)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
===Solution 1===
+
===Solution 1 (Alcumus)===
−
By [[Vieta's Formulas]], we know that <math>a</math> is the sum of the three roots of the polynomial <math>x^3-ax^2+bx-2010</math>. Again Vieta's Formulas tell us that <math>2010</math> is the product of the three integer roots. Also, <math>2010</math> factors into <math>2\cdot3\cdot5\cdot67</math>. But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with three roots. To minimize <math>a</math>, <math>2</math> and <math>3</math> should be multiplied, which means <math>a</math> will be <math>6+5+67=78</math> and the answer is <math>\boxed{\textbf{(A)78}}</math>.
+
By [[Vieta's Formulas]], we know that <math>a</math> is the sum of the three roots of the polynomial <math>x^3-ax^2+bx-2010</math>. Again Vieta's Formulas tell us that <math>2010</math> is the product of the three integer roots. Also, <math>2010</math> factors into <math>2\cdot3\cdot5\cdot67</math>. But, since the polynomial has only three roots, two of the four prime factors must be multiplied so that we are left with three roots. To minimize <math>a</math>, <math>2</math> and <math>3</math> should be multiplied, which means <math>a</math> will be <math>6+5+67=78</math> and the answer is <math>\boxed{\textbf{(A)78}}</math>.
 
~JimPickens
 
~JimPickens
 +
 +
 +
 +
 +
 +
Note:
 +
 +
If you are feeling unconfident about <math>78</math>, you can try to expand the expression <math>(x-5)(x-6)(x-67)</math> which has roots <math>5</math>, <math>6</math>, and <math>67</math>.
 +
<cmath>(x-5)(x-6)(x-67)=(x^{2}-11x+30)(x-67)</cmath>
 +
<cmath>=x\left(x^{2}-11x+30\right)-67\left(x^{2}-11x+30\right)=x^{3}-11x^{2}+30x-67x^{2}+737x-2010=x^{3}-78x^{2}+767x-2010</cmath>
 +
 +
As we can see, <math>a=78</math>, and since this is the least answer choice, we can be confident that the right option is <math>\boxed{\textbf{(A) } 78}</math>.
 +
 +
~JH. L
  
 
===Solution 2===
 
===Solution 2===
−
We can expand <math>(x+a)(x+b)(x+c)</math> as <math>(x^2+ax+bx+ab)(x+c)</math>
+
We can expand <math>(x+a)(x+b)(x+c)</math> as \begin{align*}(x+a)(x+b)(x+c) &= (x^2+ax+bx+ab)(x+c) \\
 +
&= x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc \\
 +
&= x^3+x^2(a+b+c)+x(ab+ac+bc)+abc \end{align*}
  
−
<math>(x^2+ax+bx+ab)(x+c)=x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc=x^3+x^2(a+b+c)+x(ab+ac+bc)+abc</math>
+
We do not care about <math>+bx</math> in this case, because we are only looking for <math>a</math>.  We know that the constant term is <math>-2010=-(2\cdot 3\cdot 5\cdot 67)</math>
−
 
 
−
We do not care about <math>+bx</math> in this case, because we are only looking for a.  We know that the constant term is <math>-2010=-(2\cdot 3\cdot 5\cdot 67)</math>
 
 
We are trying to minimize a, such that we have <math>-ax^2</math>
 
We are trying to minimize a, such that we have <math>-ax^2</math>
 
Since we have three positive solutions, we have <math>(x-a)(x-b)(x-c)</math> as our factors.  We have to combine two of the factors of <math>2\cdot 3\cdot 5\cdot 67</math>, and then sum up the <math>3</math> resulting factors.  Since we are minimizing, we choose <math>2</math> and <math>3</math> to combine together.  We get <math>(x-6)(x-5)(x-67)</math> which gives us a coefficient of <math>x^2</math> of <math>-6-5-67=-78</math>
 
Since we have three positive solutions, we have <math>(x-a)(x-b)(x-c)</math> as our factors.  We have to combine two of the factors of <math>2\cdot 3\cdot 5\cdot 67</math>, and then sum up the <math>3</math> resulting factors.  Since we are minimizing, we choose <math>2</math> and <math>3</math> to combine together.  We get <math>(x-6)(x-5)(x-67)</math> which gives us a coefficient of <math>x^2</math> of <math>-6-5-67=-78</math>
 
Therefore <math>-a=-78</math> or <math>a=\boxed{\textbf{(A)}78}</math>
 
Therefore <math>-a=-78</math> or <math>a=\boxed{\textbf{(A)}78}</math>
  
−
==Video Solution==
+
===Solution 3===
 +
We want the polynomial <math>x^3-ax^2+bx-2010</math> to have POSITIVE integer roots. That means we want to factor it in to the form <math>(x-a)(x-b)(x-c).</math> We therefore want the prime factorization for <math>2010</math>. The prime factorization of <math>2010</math> is <math>2 \cdot 3 \cdot 5 \cdot 67</math>. We want the smallest difference of the <math>3</math> roots since by [[Vieta's formulas]], <math>a</math> is the sum of the <math>3</math> roots.
 +
 
 +
We proceed to factorize it in to <math>(x-5)(x-6)(x-67)</math>. Therefore, our answer is <math>5+6+67</math> = <math>\boxed{\textbf{(A)78}}</math>.
 +
 
 +
~Arcticturn
 +
 
 +
 
 +
Suggestion for the author: The variables <math>a</math> and <math>b</math> are already defined as coefficients of the cubic polynomial. So, consider using different variables when describing how we want to factor this polynomial in the second sentence. Thanks!
 +
 
 +
~Jwarner
 +
 
 +
==Notes==
 +
We can check <math>5 \cdot 6 \cdot 67</math> has the smallest possible sum because of the following reasons:
 +
 
 +
<math>1)</math>: We don't want to multiply <math>67</math> by anything since that would make the sum of the roots too big.
 +
 
 +
<math>2)</math>: The smallest number should multiply since that would make the numbers optimally small. Therefore, we want <math>2</math> times <math>3</math>.
 +
 
 +
[[Vieta's Formulas]]
 +
 
 +
~Arcticturn
 +
 
 +
==Video Solution 1==
 
https://youtu.be/LCx0go2BXiY
 
https://youtu.be/LCx0go2BXiY
  
 
~IceMatrix
 
~IceMatrix
 +
 +
== Video Solution 2 ==
 +
https://youtu.be/3dfbWzOfJAI?t=2352
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2010|ab=A|num-b=20|num-a=22}}
 
{{AMC10 box|year=2010|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:41, 8 September 2024

Problem

The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$?

$\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118$

Solution

Solution 1 (Alcumus)

By Vieta's Formulas, we know that $a$ is the sum of the three roots of the polynomial $x^3-ax^2+bx-2010$. Again Vieta's Formulas tell us that $2010$ is the product of the three integer roots. Also, $2010$ factors into $2\cdot3\cdot5\cdot67$. But, since the polynomial has only three roots, two of the four prime factors must be multiplied so that we are left with three roots. To minimize $a$, $2$ and $3$ should be multiplied, which means $a$ will be $6+5+67=78$ and the answer is $\boxed{\textbf{(A)78}}$. ~JimPickens



Note:

If you are feeling unconfident about $78$, you can try to expand the expression $(x-5)(x-6)(x-67)$ which has roots $5$, $6$, and $67$. \[(x-5)(x-6)(x-67)=(x^{2}-11x+30)(x-67)\] \[=x\left(x^{2}-11x+30\right)-67\left(x^{2}-11x+30\right)=x^{3}-11x^{2}+30x-67x^{2}+737x-2010=x^{3}-78x^{2}+767x-2010\]

As we can see, $a=78$, and since this is the least answer choice, we can be confident that the right option is $\boxed{\textbf{(A) } 78}$.

~JH. L

Solution 2

We can expand $(x+a)(x+b)(x+c)$ as \begin{align*}(x+a)(x+b)(x+c) &= (x^2+ax+bx+ab)(x+c) \\ &= x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc \\ &= x^3+x^2(a+b+c)+x(ab+ac+bc)+abc \end{align*}

We do not care about $+bx$ in this case, because we are only looking for $a$. We know that the constant term is $-2010=-(2\cdot 3\cdot 5\cdot 67)$ We are trying to minimize a, such that we have $-ax^2$ Since we have three positive solutions, we have $(x-a)(x-b)(x-c)$ as our factors. We have to combine two of the factors of $2\cdot 3\cdot 5\cdot 67$, and then sum up the $3$ resulting factors. Since we are minimizing, we choose $2$ and $3$ to combine together. We get $(x-6)(x-5)(x-67)$ which gives us a coefficient of $x^2$ of $-6-5-67=-78$ Therefore $-a=-78$ or $a=\boxed{\textbf{(A)}78}$

Solution 3

We want the polynomial $x^3-ax^2+bx-2010$ to have POSITIVE integer roots. That means we want to factor it in to the form $(x-a)(x-b)(x-c).$ We therefore want the prime factorization for $2010$. The prime factorization of $2010$ is $2 \cdot 3 \cdot 5 \cdot 67$. We want the smallest difference of the $3$ roots since by Vieta's formulas, $a$ is the sum of the $3$ roots.

We proceed to factorize it in to $(x-5)(x-6)(x-67)$. Therefore, our answer is $5+6+67$ = $\boxed{\textbf{(A)78}}$.

~Arcticturn


Suggestion for the author: The variables $a$ and $b$ are already defined as coefficients of the cubic polynomial. So, consider using different variables when describing how we want to factor this polynomial in the second sentence. Thanks!

~Jwarner

Notes

We can check $5 \cdot 6 \cdot 67$ has the smallest possible sum because of the following reasons:

$1)$: We don't want to multiply $67$ by anything since that would make the sum of the roots too big.

$2)$: The smallest number should multiply since that would make the numbers optimally small. Therefore, we want $2$ times $3$.

Vieta's Formulas

~Arcticturn

Video Solution 1

https://youtu.be/LCx0go2BXiY

~IceMatrix

Video Solution 2

https://youtu.be/3dfbWzOfJAI?t=2352

~ pi_is_3.14

See Also

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_21&oldid=226797"