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

(Created page with "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...")
 
Line 5: Line 5:
 
<math>IV</math> is true because <math>2+3=5</math> is prime.  
 
<math>IV</math> is true because <math>2+3=5</math> is prime.  
 
Thus, the answer is <math>\textbf{(E)}</math>.
 
Thus, the answer is <math>\textbf{(E)}</math>.
 +
-brainiacmaniac31

Revision as of 22:38, 6 January 2020

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

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