Difference between revisions of "2010 AMC 10A Problems/Problem 21"
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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= |
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=A|num-b=20|num-a=22}} | {{AMC10 box|year=2010|ab=A|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:50, 12 June 2019
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 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 
.
~JimPickens
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 
==Solution 3
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
