Difference between revisions of "2017 USAJMO Problems/Problem 4"

(Problem)
(Solution)
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==Solution==
 
==Solution==
  
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Yes. Take <math>a=b=c=1</math> then <math>11 \mid -2013 = 1^2 + 1^2 + 1^2 + 1\times1\times1 - 2017</math>.
 
{{MAA Notice}}
 
{{MAA Notice}}
  
 
==See also==
 
==See also==
 
{{USAJMO newbox|year=2017|num-b=3|num-a=5}}
 
{{USAJMO newbox|year=2017|num-b=3|num-a=5}}

Revision as of 22:30, 9 March 2019

Problem

Are there any triples $(a,b,c)$ of positive integers such that $(a-2)(b-2)(c-2) + 12$ is prime that properly divides the positive number $a^2 + b^2 + c^2 + abc - 2017$?

Solution

Yes. Take $a=b=c=1$ then $11 \mid -2013 = 1^2 + 1^2 + 1^2 + 1\times1\times1 - 2017$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

2017 USAJMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6
All USAJMO Problems and Solutions