Question 569241
It is generally best to avoid posting this question until the day after the AMC10/12A exams (yes, I took this exam as well). But since it's pretty late in the day, I'll show the solution. Make sure not to publicize it to anyone else.


Adding both equations (I should have done this myself during the exam...I tried adding three times the first equation to the second equation and got nowhere).


*[tex \LARGE 2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14]


This factors to


*[tex \LARGE (a-b)^2 + (a-c)^2 + (b-c)^2 = 14]


Essentially we have three square numbers adding to 14, so the only possibility is {9,4,1}. Since a and c are the furthest apart, we have


*[tex \LARGE (a-c)^2 = 9 \Rightarrow a-c = 3]


So our three numbers are either {c,c+1,c+3} or {c,c+2,c+3}. Suppose the numbers are {c,c+1,c+3}. Plugging into the first equation we have


*[tex \LARGE (c+3)^2 - (c+1)^2 - c^2 + (c+1)(c+3) = 2011 \Rightarrow c = 250] (this is actually a linear equation)


If c = 250, then a = 253 and we are done. For sake of completeness we can try the other case:


*[tex \LARGE (c+3)^2 - (c+2)^2 - c^2 + (c+2)(c+3) = 2011], however if you solved this, c would not be an integer. Therefore a = 253.