Forcing yourself to only rely on the solution strategies taught by school makes you very bad at problem solving.
Also ny solution wasn't brute force or trial and error, there are very few possible sequences of 11 numbers around 201, so you are guaranteed to find the solution quickly.
Perhaps the elementary school math contest permits the use of a calculator? That would make it very quick for sixth graders to solve it using your method.
Assume the middle of the sequence is 201 since 2012 / 10 ~= 201. Then the sum of the 11 numbers between 195 and 206 would be 11 * 201 = 2211. Subtract the sum of the values we didn't remove to get 2211 - 2012 = 199. 199 is between 195 and 206 proving that it is a valid answer.
Edit: Also I did my masters in pure math, doesn't mean that I cannot use common logic without the formalism.
Try the same trick with this modified problem:
Fifty consecutive positive integers are written on a board. Maria erases one of the numbers. If the sum of the remaining numbers is 2009, what number did Maria erase?
You'll find that it does not work. The trick you did relies on the specific problem as stated. It is not true in general.
2009 / 49 = 41. 41 * 50 = 2050.
2050 is not the sum of the numbers 17..66, 2075 is, and subtracting 2009 from 2050 just gets you 41 again, which is not the correct answer.
I think a school student can do this math manually. The hardest part is the sum of the first N(in this case 10) numbers and then finding the divisor. Yes there might be a better solution.
But I think a kid would think up a brute force algorithm like this and it is easier for small N. The computation is only in O(N) complexity haha. And if an answer exists it will always be found - so it is a real solution. Might be the first algorithm you write if you are tasked to do this with software.
In this case, let's consider the sequence of 50 consecutive integers 16, 17, ..., 40, 41, ..., 65 where 40.5 is in the middle and the sum is 40.5 * 50 = 2025. Subtracting 2009 we get a supposed erased value of 2025 - 2009 = 16 and this indeed correspond to a valid solution:
16, 17, ..., 65 => 17, ..., 65
We also could've considered the sequence 17, ..., 41, 42, ..., 65, 66 centered at 41.5 and this would've led to an erased value of 41.5 * 50 - 2009 = 66 which is another solution:
17, ..., 65, 66 => 17, ..., 65
For instance, if you want to find the global minimum of some function on the integers, you might realize you can find the roots of the derivative and test only the neighboring integers. If you want to divide 144359305 by 241, you're going to start by guessing how many times 241 goes into 1443, and then multiply to check whether that's right. If you want to compute floor(29489238428^(1/3)) by hand, you're better off smartly guessing and checking than trying to figure out a digit-by-digit cube root algorithm on the spot.
This sort of puzzling through play is good for solving math contest problems that have been specifically designed for the purpose, and little else.
Example: The Josephus Problem - Numberphile https://youtube.com/watch?v=uCsD3ZGzMgE&t=86
The eleven numbers are 196,197,..., 206, out of which 199 was removed.
I began with n being the largest of the sum so that 11n - 55 = 2012 + k. I solved for k in terms of n and then used an inequality to find n which gave me k.
Now I see that the method I used relies on math beyond elementary school, which makes the problem a bit more complicated than it ought to be. On the other hand, the elementary school only method relies on the fact that you have 10 numbers remaining and so you can easily see the remainder. It wouldn’t generalize as well, especially in cases where there may be multiple solutions and you’re asked to find all of them.
10*n+sum_conseq=2012
therefore erased number is n+3 because (2012-55-3) divisible by 10 -> n=196.
I solved the problem for n, the largest number in the sequence of eleven, and a, the number Maria erases, by the following:
n - 11 < a < n + 1
11(n + n - 10)/2 = 2012 + a
I’m interested to know what other methods may be used to solve it (apart from brute force).
This is a very challenging problem. Without a calculator (or computer), it took me a system of an equation with an inequality in 2 variables, using the fact that the solutions are integers, to solve it.
I would be very surprised (and excited) to meet an 8th grader (let alone a 6th) who could solve this problem without any external help the way I did. I would not expect an elementary school student to have this kind of algebra knowledge. Most elementary school students around here don’t even know what algebra is.