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

(Solution 2)
(Solution 2)
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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>
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==Solution 3==
  
 
==Video Solution 1==
 
==Video Solution 1==

Revision as of 19:16, 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

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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