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Difference between revisions of "1983 AHSME Problems/Problem 3"

(Created page with "==Problem 3== Three primes <math>p,q</math>, and <math>r</math> satisfy <math>p+q = r</math> and <math>1 < p < q</math>. Then <math>p</math> equals <math>\textbf{(A)}\ 2\qqu...")
 
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<math>\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 17</math>
 
<math>\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 17</math>
 
==Solution==
 
==Solution==
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We are given that <math>p,q</math> and <math>r</math> are primes. In order to sum two another prime, either <math>p</math> or <math>q</math> has to be even, because the sum of an odd and an even is odd. The only odd prime is <math>2</math>, and it is also the smallest prime, so therefore, the answer is <math>\fbox{\textbf{(A)}2}</math>
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We are given that <math>p,q</math> and <math>r</math> are primes. In order for <math>p</math> and <math>q</math> to sum to another prime, either <math>p</math> or <math>q</math> has to be even, because the sum of two odd numbers would be even, and the only even prime is <math>2</math> (but <math>p + q = 2</math> would have, as the only solution in positive integers, <math>p = q = 1</math>, and <math>1</math> is not prime). Thus, with one of either <math>p</math> or <math>q</math> being even, either <math>p</math> or <math>q</math> must be <math>2</math>, and as <math>p < q</math>, we deduce <math>p = 2</math> (as <math>2</math> is the smallest prime). This means the answer is <math>\boxed{\textbf{(A)}\ 2}</math>.
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==See Also==
 
==See Also==
 
{{AHSME box|year=1983|num-b=3|num-a=4}}
 
{{AHSME box|year=1983|num-b=3|num-a=4}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:41, 19 February 2019

Problem 3

Three primes $p,q$, and $r$ satisfy $p+q = r$ and $1 < p < q$. Then $p$ equals

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 17$

Solution

We are given that $p,q$ and $r$ are primes. In order for $p$ and $q$ to sum to another prime, either $p$ or $q$ has to be even, because the sum of two odd numbers would be even, and the only even prime is $2$ (but $p + q = 2$ would have, as the only solution in positive integers, $p = q = 1$, and $1$ is not prime). Thus, with one of either $p$ or $q$ being even, either $p$ or $q$ must be $2$, and as $p < q$, we deduce $p = 2$ (as $2$ is the smallest prime). This means the answer is $\boxed{\textbf{(A)}\ 2}$.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 4
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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