
Math Events Math Events, Competitions, Meetups  Local, Regional, State, National, International 
 LinkBack  Thread Tools  Display Modes 
May 23rd, 2014, 09:42 PM  #1 
Member Joined: Nov 2013 Posts: 47 Thanks: 4  Math Competition Problem
Any clues on how to solve this? I'm stuck 
May 24th, 2014, 02:29 AM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,164 Thanks: 172  Here's how i did it. $5^5 = 3125 \ $ so for positive integer $ \ a \ $ , $ \ 1 \ \le \ a \ \le \ 4 $ We check a maximum of four values of $a$ $a = 4$ $$4^5 + b^2 + c^2 = 2012$$ $$b^2 + c^2 = 988$$ $$\sqrt{988} < 32 $$ $$ \sqrt{ \frac{988}{2}} > 22 $$ So WLOG , we test $ \ 23 \ \le \ b \ \le \ 31 $ Nine values , none of them work. $a = 3$ $$ 3^5 + b^2 + c^2 = 2012$$ $$b^2 + c^2 = 1769$$ $$ \sqrt{1769} < 43$$ $$ \sqrt{\frac{1769}{2}} > 29 $$ So WLOG , we test $ \ 30 \ \le \ b \ \le \ 42$ Thirteen values , $37^2 + 20^2 = 1769$ works AND $37  20 = 17$ is prime We can stop , no need to test any other values of $a$ since the conditions are satisfied. THEREFORE $$a = 3 $$ $$b = 37$$ $$c = 20$$ $$a + b + c = 60$$ Coice E. Note: $3^5 + 40^2 + 13^2 = 2012$ but $40  13 = 27$ is NOT prime so choice D is disqualified. Last edited by agentredlum; May 24th, 2014 at 02:52 AM. Reason: The first edit was incorrect. It could be corrected but leads to more complicated and more tedious calculations. 
May 24th, 2014, 06:34 AM  #3 
Member Joined: Nov 2013 Posts: 47 Thanks: 4 
Thanks! My approach was similar and I was able to deduce that a had to be less than 5, I was just trying to come up with a more direct approach rather than resorting to a sort of trial and error method. But hey, at least it gets us the answer. Thanks again! 
May 24th, 2014, 07:56 AM  #4 
Math Team Joined: Apr 2010 Posts: 2,770 Thanks: 356 
For a = 4 we can also work as follows: If a = 4 then 4^5 + b^2 + c^2 = 2012. b^2 + c^2 = 988 if $\displaystyle \text{sign}(c) \ne \text{sign}(b)$ then b^2 + c^2 is odd. Since 988 is even, that's not an option. If the signs are equal, then b  c is even, so, since it's prime, it's two. Now, we just need to check that there are no solutions to b^2 + c^2 = (c+2)^2 + c^2 = 988. 

Tags 
competition, math, problem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Math Competition Problem (I believe trig is involved)  Orlando895  Trigonometry  4  April 1st, 2014 10:54 AM 
Math Competition Problem  icemanfan  Algebra  3  March 15th, 2012 06:45 PM 
Interactive Math Competition with Prizes  PaidGrade  New Users  0  March 14th, 2012 11:50 AM 
Another question on the math competition  asddsa  Algebra  0  November 16th, 2011 04:06 PM 
Mental Math world competition  pgato  Academic Guidance  2  June 7th, 2010 11:35 AM 