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

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==Problem==
 
==Problem==
The polynomial <math>x^3-ax^2+bx-2010</math> has three positive integer zeros. What is the smallest possible value of <math>a</math>?
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The polynomial <math>x^3-ax^2+bx-2010</math> has three positive integer roots. What is the smallest possible value of <math>a</math>?
  
 
<math>\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118</math>
 
<math>\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118</math>
==Solution 1==
 
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>.
 
Also, 2010 factors into <math>2*3*5*67</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)}}</math>.
 
  
==Solution 2==
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==Solution==
The polynomial <math> x^{3}-ax^{2}+bx-2010 </math> as three positive integer zeros. What is the smallest possible value of <math>a</math>? [/quote]
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===Solution 1===
 +
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
 +
 
 +
===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 <math>(x^2+ax+bx+ab)(x+c)</math>
 +
 
<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>
 
<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 a.  We know that the constant term is <math>-2010=-(2*3*5*67)</math>
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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 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*3*5*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>
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Therefore <math>-a=-78</math> or <math>a=\boxed{\textbf{(A)}78}</math>
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 +
==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> root
 +
 
 +
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
 +
 
 +
==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
 +
 
 +
~IceMatrix
 +
 
 +
== Video Solution 2 ==
 +
https://youtu.be/3dfbWzOfJAI?t=2352
 +
 
 +
~ pi_is_3.14
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2010|ab=A|num-b=20|num-a=22}}
 +
{{MAA Notice}}

Revision as of 20:29, 24 October 2021

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

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

Solution 2

We can expand $(x+a)(x+b)(x+c)$ as $(x^2+ax+bx+ab)(x+c)$

$(x^2+ax+bx+ab)(x+c)=x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc=x^3+x^2(a+b+c)+x(ab+ac+bc)+abc$

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 $3$ root

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

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 (ProblemsAnswer KeyResources)
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

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