Difference between revisions of "2010 AMC 10A Problems/Problem 21"
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==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). We therefore want the prime factorization for < | + | 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>. Since we want 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> numbers since by [[Vieta's formulas]], <math>a</math> is the sum of the <math>3</math> roots. |
− | We proceed to factorize it in to < | + | 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>. |
==Video Solution 1== | ==Video Solution 1== |
Revision as of 19:23, 24 October 2021
Contents
[hide]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 tell us that is the product of the three integer roots. Also, factors into . 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 , and should be multiplied, which means will be and the answer is . ~JimPickens
Solution 2
We can expand as
We do not care about in this case, because we are only looking for . 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
Solution 3
We want the polynomial to have POSITIVE integer roots. That means we want to factor it in to the form We therefore want the prime factorization for . Since we want the prime factorization of is , we want the smallest difference of the numbers since by Vieta's formulas, is the sum of the roots.
We proceed to factorize it in to . Therefore, our answer is = .
Video Solution 1
~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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