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May 23rd, 2014, 09:42 PM   #1
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Math Competition Problem

Any clues on how to solve this? I'm stuck Attached Images Screenshot 2014-05-24 at 1.36.56 AM.jpg (10.9 KB, 19 views) May 24th, 2014, 02:29 AM   #2
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Quote:
 Originally Posted by Orlando895 Any clues on how to solve this? I'm stuck 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! Thanks from agentredlum May 24th, 2014, 07:56 AM #4 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 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. Thanks from agentredlum Tags competition, math, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Orlando895 Trigonometry 4 April 1st, 2014 10:54 AM icemanfan Algebra 3 March 15th, 2012 06:45 PM PaidGrade New Users 0 March 14th, 2012 11:50 AM asddsa Algebra 0 November 16th, 2011 03:06 PM pgato Academic Guidance 2 June 7th, 2010 11:35 AM

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