# 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}. | + | III.\ p^2-q\ \text{is prime}. \\ |

− | IV.\ p+q \text{is prime} | + | 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 | + | \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 | + | ==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 | ||

+ | |||

+ | ==See Also== | ||

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

+ | {{MAA Notice}} |

## Revision as of 17:48, 19 January 2021

## Problem

If and are primes and has distinct positive integral roots, then which of the following statements are true?

## Solution

Since the roots are both positive integers, we can say that since only has divisors. Thus, the roots are and and . The only two primes which differ by are so and . is true because . is true because one of the roots is which is prime. is true because is prime. is true because is prime. Thus, the answer is . -brainiacmaniac31

## See Also

{{AHSME box|year=1975|num-b=21|num-a=23} The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.