Difference between revisions of "1975 AHSME Problems/Problem 22"

 
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I.\ \text{The difference of the roots is odd.} \\
 
I.\ \text{The difference of the roots is odd.} \\
 
II.\ \text{At least one root is prime.} \\
 
II.\ \text{At least one root is prime.} \\
III.\ p^2-q\ \text{is prime}.
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III.\ p^2-q\ \text{is prime}. \\
IV.\ p+q \text{is prime}
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IV.\ p+q\ \text{is prime}
 
</math>
 
</math>
  
 
<math>
 
<math>
 +
\\
 
\textbf{(A)}\ I\ \text{only} \qquad
 
\textbf{(A)}\ I\ \text{only} \qquad
 
\textbf{(B)}\ II\ \text{only} \qquad
 
\textbf{(B)}\ II\ \text{only} \qquad
 
\textbf{(C)}\ II\ \text{and}\ III\ \text{only} \\
 
\textbf{(C)}\ II\ \text{and}\ III\ \text{only} \\
\textbf{(D)}\ I, II, \text{and}\ IV \text{only} \qquad
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\textbf{(D)}\ I, II, \text{and}\ IV\ \text{only}\ \qquad
 
\textbf{(E)}\ \text{All are true.}
 
\textbf{(E)}\ \text{All are true.}
 
 
</math>
 
</math>
  
==Solution
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==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>.  
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Thus, the answer is <math>\textbf{(E)}</math>.
 
Thus, the answer is <math>\textbf{(E)}</math>.
 
-brainiacmaniac31
 
-brainiacmaniac31
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 +
==See Also==
 +
{{AHSME box|year=1975|num-b=21|num-a=23}}
 +
{{MAA Notice}}

Latest revision as of 16:48, 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

See Also

1975 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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 26 27 28 29 30
All AHSME Problems and Solutions

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