Difference between revisions of "2010 AMC 10A Problems/Problem 21"
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[[Vieta's Formulas]] | [[Vieta's Formulas]] | ||
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+ | ~Arcticturn | ||
==Video Solution 1== | ==Video Solution 1== |
Revision as of 20:29, 24 October 2021
Contents
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
. The prime factorization of
is
. We want the smallest difference of the
roots since by Vieta's formulas,
is the sum of the
root
We proceed to factorize it in to . Therefore, our answer is
=
.
~Arcticturn
Notes
We can check has the smallest possible sum because of the following reasons:
: We don't want to multiply
by anything since that would make the sum of the roots too big.
: The smallest number should multiply since that would make the numbers optimally small. Therefore, we want
times
.
~Arcticturn
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.