Difference between revisions of "2010 AMC 10A Problems/Problem 21"
Elephant200 (talk | contribs) m (→Solution 2) |
|||
(30 intermediate revisions by 19 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | The polynomial <math>x^3-ax^2+bx-2010</math> has three positive integer | + | The polynomial <math>x^3-ax^2+bx-2010</math> has three positive integer roots. What is the smallest possible value of <math>a</math>? |
<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== | ==Solution== | ||
− | ===Solution 1=== | + | ===Solution 1 (Alcumus)=== |
− | 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 tell 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 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 <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)78}}</math>. |
− | Also, 2010 factors into <math>2\cdot3\cdot5\cdot67</math>. But, since | + | ~JimPickens |
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | Note: | ||
+ | |||
+ | If you are feeling unconfident about <math>78</math>, you can try to expand the expression <math>(x-5)(x-6)(x-67)</math> which has roots <math>5</math>, <math>6</math>, and <math>67</math>. | ||
+ | <cmath>(x-5)(x-6)(x-67)=(x^{2}-11x+30)(x-67)</cmath> | ||
+ | <cmath>=x\left(x^{2}-11x+30\right)-67\left(x^{2}-11x+30\right)=x^{3}-11x^{2}+30x-67x^{2}+737x-2010=x^{3}-78x^{2}+767x-2010</cmath> | ||
+ | |||
+ | As we can see, <math>a=78</math>, and since this is the least answer choice, we can be confident that the right option is <math>\boxed{\textbf{(A) } 78}</math>. | ||
+ | |||
+ | ~JH. L | ||
===Solution 2=== | ===Solution 2=== | ||
− | We can expand <math>(x+a)(x+b)(x+c)</math> as | + | We can expand <math>(x+a)(x+b)(x+c)</math> as \begin{align*}(x+a)(x+b)(x+c) &= (x^2+ax+bx+ab)(x+c) \ |
+ | &= x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc \ | ||
+ | &= x^3+x^2(a+b+c)+x(ab+ac+bc)+abc \end{align*} | ||
+ | |||
+ | We do not care about <math>+bx</math> in this case, because we are only looking for <math>a</math>. We know that the constant term is <math>-2010=-(2\cdot 3\cdot 5\cdot 67)</math> | ||
+ | We are trying to minimize a, such that we have <math>-ax^2</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\cdot 3\cdot 5\cdot 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> | ||
− | <math> | + | ===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).</math> We therefore want the prime factorization for <math>2010</math>. 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> roots since by [[Vieta's formulas]], <math>a</math> is the sum of the <math>3</math> roots. | ||
− | We | + | 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>. |
− | + | ||
− | + | ~Arcticturn | |
− | + | ||
+ | |||
+ | Suggestion for the author: The variables <math>a</math> and <math>b</math> are already defined as coefficients of the cubic polynomial. So, consider using different variables when describing how we want to factor this polynomial in the second sentence. Thanks! | ||
+ | |||
+ | ~Jwarner | ||
+ | |||
+ | ==Notes== | ||
+ | We can check <math>5 \cdot 6 \cdot 67</math> has the smallest possible sum because of the following reasons: | ||
+ | |||
+ | <math>1)</math>: We don't want to multiply <math>67</math> by anything since that would make the sum of the roots too big. | ||
+ | |||
+ | <math>2)</math>: The smallest number should multiply since that would make the numbers optimally small. Therefore, we want <math>2</math> times <math>3</math>. | ||
+ | |||
+ | [[Vieta's Formulas]] | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | https://youtu.be/LCx0go2BXiY | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | == Video Solution 2 == | ||
+ | https://youtu.be/3dfbWzOfJAI?t=2352 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==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}} |
Latest revision as of 19:41, 8 September 2024
Contents
[hide]Problem
The polynomial has three positive integer roots. What is the smallest possible value of
?
Solution
Solution 1 (Alcumus)
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
Note:
If you are feeling unconfident about , you can try to expand the expression
which has roots
,
, and
.
As we can see, , and since this is the least answer choice, we can be confident that the right option is
.
~JH. L
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
roots.
We proceed to factorize it in to . Therefore, our answer is
=
.
~Arcticturn
Suggestion for the author: The variables and
are already defined as coefficients of the cubic polynomial. So, consider using different variables when describing how we want to factor this polynomial in the second sentence. Thanks!
~Jwarner
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.