Difference between revisions of "2010 AMC 10A Problems/Problem 21"
(→Solution) |
(fixed formatting issues - added box) |
||
Line 3: | Line 3: | ||
<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== | + | ==Solution== |
+ | ===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>. | 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>. | 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== | + | ===Solution 2=== |
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] | 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] | ||
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> | ||
Line 16: | Line 17: | ||
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*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> | ||
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> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2010|ab=A|num-b=20|num-a=22}} |
Revision as of 13:38, 6 June 2011
Contents
[hide]Problem
The polynomial has three positive integer zeros. What is the smallest possible value of ?
Solution
Solution 1
By Vieta's Formulas, we know that is the sum of the three roots of the polynomial . Also, 2010 factors into . 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 , and should be multiplied, which means will be and the answer is .
Solution 2
The polynomial as three positive integer zeros. What is the smallest possible value of ? [/quote] We can expand as
We do not care about in this case, because we are only looking for a. We know that the constant term is We are trying to minimize a, such that we have Since we have three positive solutions, we have as our factors. We have to combine two of the factors of , and then sum up the resulting factors. Since we are minimizing, we choose and to combine together. We get which gives us a coefficient of of Therefore or
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 |