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}} | ||
Latest revision as of 16:48, 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
| 1975 AHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
