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

 
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==See Also==
 
==See Also==
{{AHSME box|year=1975|num-b=21|num-a=23}
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{{AHSME box|year=1975|num-b=21|num-a=23}}
 
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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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