Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1975 AHSME Problems/Problem 22"
Line 1: Line 1:
 +
==Problem==
 +
If <math>p</math> and <math>q</math> are primes and <math>x^2-px+q=0</math> has distinct positive integral roots, then which of the following statements are true?
 +
 +
<math>
 +
I.\ \text{The difference of the roots is odd.} \\
 +
II.\ \text{At least one root is prime.} \\
 +
III.\ p^2-q\ \text{is prime}.
 +
IV.\ p+q \text{is prime}
 +
</math>
 +
 +
<math>
 +
\textbf{(A)}\ I\ \text{only} \qquad
 +
\textbf{(B)}\ II\ \text{only} \qquad
 +
\textbf{(C)}\ II\ \text{and}\ III\ \text{only} \\
 +
\textbf{(D)}\ I, II, \text{and}\ IV \text{only} \qquad
 +
\textbf{(E)}\ \text{All are true.}
 +
 +
</math>
 +
 +
==Solution
 +
 
Since the roots are both positive integers, we can say that <math>x^2-px+q=(x-1)(x-q)</math> since <math>q</math> only has <math>2</math> divisors. Thus, the roots are <math>1</math> and <math>q</math> and <math>p=q+1</math>. The only two primes which differ by <math>1</math> are <math>2,3</math> so <math>p=3</math> and <math>q=2</math>.  
 
Since the roots are both positive integers, we can say that <math>x^2-px+q=(x-1)(x-q)</math> since <math>q</math> only has <math>2</math> divisors. Thus, the roots are <math>1</math> and <math>q</math> and <math>p=q+1</math>. The only two primes which differ by <math>1</math> are <math>2,3</math> so <math>p=3</math> and <math>q=2</math>.  
 
<math>I</math> is true because <math>3-2=1</math>.
 
<math>I</math> is true because <math>3-2=1</math>.

Revision as of 17:44, 19 January 2021

Problem

If $p$ and $q$ are primes and $x^2-px+q=0$ has distinct positive integral roots, then which of the following statements are true?

$I.\ \text{The difference of the roots is odd.} \\ II.\ \text{At least one root is prime.} \\ III.\ p^2-q\ \text{is prime}. IV.\ p+q \text{is prime}$

$\textbf{(A)}\ I\ \text{only} \qquad \textbf{(B)}\ II\ \text{only} \qquad \textbf{(C)}\ II\ \text{and}\ III\ \text{only} \\ \textbf{(D)}\ I, II, \text{and}\ IV \text{only} \qquad \textbf{(E)}\ \text{All are true.}$

==Solution

Since the roots are both positive integers, we can say that $x^2-px+q=(x-1)(x-q)$ since $q$ only has $2$ divisors. Thus, the roots are $1$ and $q$ and $p=q+1$. The only two primes which differ by $1$ are $2,3$ so $p=3$ and $q=2$. $I$ is true because $3-2=1$. $II$ is true because one of the roots is $2$ which is prime. $III$ is true because $3^2-2=7$ is prime. $IV$ is true because $2+3=5$ is prime. Thus, the answer is $\textbf{(E)}$. -brainiacmaniac31

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1975_AHSME_Problems/Problem_22&oldid=142834"