Difference between revisions of "2002 AMC 12A Problems/Problem 12"
Mathboy282 (talk | contribs) (→Solution 3) |
Mathmonkey12 (talk | contribs) (→Solution) |
||
(5 intermediate revisions by 4 users not shown) | |||
Line 12: | Line 12: | ||
Consider a general quadratic with the coefficient of <math>x^2</math> being <math>1</math> and the roots being <math>r</math> and <math>s</math>. It can be factored as <math>(x-r)(x-s)</math> which is just <math>x^2-(r+s)x+rs</math>. Thus, the sum of the roots is the negative of the coefficient of <math>x</math> and the product is the constant term. (In general, this leads to [[Vieta's Formulas]]). | Consider a general quadratic with the coefficient of <math>x^2</math> being <math>1</math> and the roots being <math>r</math> and <math>s</math>. It can be factored as <math>(x-r)(x-s)</math> which is just <math>x^2-(r+s)x+rs</math>. Thus, the sum of the roots is the negative of the coefficient of <math>x</math> and the product is the constant term. (In general, this leads to [[Vieta's Formulas]]). | ||
− | We now have that the sum of the two roots is <math>63</math> while the product is <math>k</math>. Since both roots are primes, one must be <math>2</math>, otherwise the sum would be even. That means the other root is <math>61</math> and the product must be <math>122</math>. Hence, our answer is <math>\boxed{\text{(B)}\ 1 }</math>. | + | We now have that the sum of the two roots is <math>63</math> while the product is <math>k</math>. Since both roots are primes, one must be <math>2</math>, otherwise, the sum would be even. That means the other root is <math>61</math> and the product must be <math>122</math>. Hence, our answer is <math>\boxed{\text{(B)}\ 1 }</math>. |
==Solution 2== | ==Solution 2== | ||
Line 21: | Line 21: | ||
By Vieta's you have | By Vieta's you have | ||
− | < | + | <cmath>a+b=63</cmath> |
<cmath>ab=k</cmath> | <cmath>ab=k</cmath> | ||
− | + | Since we know odd + even = odd, we must have either <math>a</math> or <math>b</math> equal to <math>2</math> and the other equal to <math>63-2=61.</math> Both of these are prime so they satisfy the restraints. Thus there is <math>\boxed{(B) 1}</math> solution. | |
== Video Solution == | == Video Solution == | ||
+ | https://youtu.be/0c7QMWYU8So | ||
+ | |||
https://youtu.be/5QdPQ3__a7I?t=130 | https://youtu.be/5QdPQ3__a7I?t=130 | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | == Note == | ||
+ | All these solutions are principled in [[Vieta's Formulas]], and can be viewed as identical. | ||
==See Also== | ==See Also== |
Revision as of 13:16, 8 October 2023
- The following problem is from both the 2002 AMC 12A #12 and 2002 AMC 10A #14, so both problems redirect to this page.
Problem
Both roots of the quadratic equation are prime numbers. The number of possible values of
is
Solution
Consider a general quadratic with the coefficient of being
and the roots being
and
. It can be factored as
which is just
. Thus, the sum of the roots is the negative of the coefficient of
and the product is the constant term. (In general, this leads to Vieta's Formulas).
We now have that the sum of the two roots is while the product is
. Since both roots are primes, one must be
, otherwise, the sum would be even. That means the other root is
and the product must be
. Hence, our answer is
.
Solution 2
Let the quadratic be factored as . Then
and
. We know that both
and
are prime, and all primes but
are odd. The sum of two odd numbers is also always even, but in this case the sum of two primes is odd, which means that one of the primes is
. This leaves only one possible value for the other root and one possible value for their product
.
Solution 3
By Vieta's you have
Since we know odd + even = odd, we must have either
or
equal to
and the other equal to
Both of these are prime so they satisfy the restraints. Thus there is
solution.
Video Solution
https://youtu.be/5QdPQ3__a7I?t=130
~ pi_is_3.14
Note
All these solutions are principled in Vieta's Formulas, and can be viewed as identical.
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 12 Problems and Solutions |
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.