Difference between revisions of "2010 AMC 10A Problems/Problem 21"
Flyingsledge (talk | contribs) m (I'm pretty sure that one of the words is a mistake.) |
AllenWang314 (talk | contribs) (→Solution 1) |
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==Solution== | ==Solution== | ||
===Solution 1=== | ===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>. Again Vieta's Formulas tells 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)}}</math>. |
− | Also, 2010 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)}}</math>. | ||
===Solution 2=== | ===Solution 2=== |
Revision as of 19:13, 30 June 2014
Problem
The polynomial has three positive integer roots. 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 . Again Vieta's Formulas tells us that is the product of the three integer roots. Also, 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
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 |
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