Difference between revisions of "1983 AHSME Problems/Problem 3"
Quantummech (talk | contribs) (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== | ||
| − | 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> | + | 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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==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}} | ||
Revision as of 07:19, 18 May 2016
Problem 3
Three primes 
, and 
satisfy 
and 
. Then 
equals
Solution
We are given that and are primes. In order to sum two another prime, either or 
has to be even, because the sum of an odd and an even is odd. The only odd prime is 
, and it is also the smallest prime, so therefore, the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
