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Difference between revisions of "2002 AMC 12A Problems/Problem 12"

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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]]).
  
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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>.
  
 
==Solution 2==
 
==Solution 2==
  
 
Let the quadratic be factored as <math>(x-r)(x-s)</math>. Then <math>-r-s=-63</math> and <math>rs=k</math>. We know that both <math>r</math> and <math>s</math> are prime, and all primes but <math>2</math> 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 <math>2</math>. This leaves only one possible value for the other root and one possible value for their product <math>k</math>.
 
Let the quadratic be factored as <math>(x-r)(x-s)</math>. Then <math>-r-s=-63</math> and <math>rs=k</math>. We know that both <math>r</math> and <math>s</math> are prime, and all primes but <math>2</math> 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 <math>2</math>. This leaves only one possible value for the other root and one possible value for their product <math>k</math>.
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 +
==Solution 3==
 +
 +
By Vieta's you have
 +
<cmath>a+b=63</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.
 +
 +
==Video Solution by Daily Dose of Math==
 +
 +
https://youtu.be/29XpsoXTXto
 +
 +
~Thesmartgreekmathdude
 +
 +
=== Note ===
 +
 +
The solution in this video may be similar to others.
  
 
==See Also==
 
==See Also==

Latest revision as of 08:55, 1 September 2024

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 $\boxed{\text{(B)}\ 1 }$.

Solution 2

Let the quadratic be factored as $(x-r)(x-s)$. Then $-r-s=-63$ and $rs=k$. We know that both $r$ and $s$ are prime, and all primes but $2$ 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 $2$. This leaves only one possible value for the other root and one possible value for their product $k$.

Solution 3

By Vieta's you have \[a+b=63\] \[ab=k\] Since we know odd + even = odd, we must have either $a$ or $b$ equal to $2$ and the other equal to $63-2=61.$ Both of these are prime so they satisfy the restraints. Thus there is $\boxed{(B) 1}$ solution.

Video Solution

https://youtu.be/0c7QMWYU8So

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.

Video Solution by Daily Dose of Math

https://youtu.be/29XpsoXTXto

~Thesmartgreekmathdude

Note

The solution in this video may be similar to others.

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

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