Earlier quoted context omitted.
You could do it without a calculator like this: 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…
199 is between 195 and 206 proving that it is a valid answer. 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 * 5…
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