Difference between revisions of "2002 AMC 12A Problems/Problem 12"
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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>. | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/5QdPQ3__a7I?t=130 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Revision as of 22:40, 17 January 2021
- The following problem is from both the 2002 AMC 12A #12 and 2002 AMC 10A #14, so both problems redirect to this page.
Contents
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 .
Video Solution
https://youtu.be/5QdPQ3__a7I?t=130
~ pi_is_3.14
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.