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>.
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== Video Solution ==
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https://youtu.be/5QdPQ3__a7I?t=130
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~ 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.


Problem

Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$

Solution

Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$. It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$. Thus, the sum of the roots is the negative of the coefficient of $x$ 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 $63$ while the product is $k$. Since both roots are primes, one must be $2$, otherwise the sum would be even. That means the other root is $61$ and the product must be $122$. Hence, our answer is $\boxed{\text{(B)}\ 1 }$.

Video Solution

https://youtu.be/5QdPQ3__a7I?t=130

~ pi_is_3.14

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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