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2010 AMC 10A Problems/Problem 21: Difference between revisions

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==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 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>.
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 \\
<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>
&= x^3+x^2(a+b+c)+x(ab+ac+bc)+abc \end{align*}


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 <math>a</math>.  We know that the constant term is <math>-2010=-(2\cdot 3\cdot 5\cdot 67)</math>
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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>


==Solution 3==
===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 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>.
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>.
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Suggestion: 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!
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
~Jwarner

Latest revision as of 23:21, 29 August 2025

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

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