Math Competition Problem

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
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.

:)
 
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Nov 2013
47
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! :D
 
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Hoempa

Math Team
Apr 2010
2,780
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.
 
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