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You wondered how a sixth grader could solve it, I showed you how. Being able to do it using formal methods is good, but my point is that sixth graders are still

You wondered how a sixth grader could solve it, I showed you how. Being able to do it using formal methods is good, but my point is that sixth graders are still pretty smart and can use more creative methods which still works very well as I showed. 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.

chongli 7y ago
I'm a pure math student and when I first found the problem I sent it to my friend who is a stats student. She solved it very quickly, but admitted to using a calculator. None of my exams permit calculators, so I've gotten used to never using them. 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.
- username90 7y ago
  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 the formalism.
  - chongli 7y ago
    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 * 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.
    - taesko 7y ago
      I solved it like this: The numbers are x, x+1..., x+10 = 11x + 55. One is erased therefore 10x + 55 - y = 2012. 10x - y = 1957. We know X is whole so we need a number between 1957 and 1967 that divides 10 and that is X. 1960 so X = 196 and therefore y = 3. Answer is X+Y = 199. 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.
    - vworldv 7y ago
      Due to a parity issue (50 here is even) we need to adjust username90's solution. In particular, a sequence of consecutive integers of even length has a half-integer in its middle. 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
      - chongli 7y ago
        Right, but we are talking about sixth-graders here. If a sixth-grader saw username90's solution and tried to apply it to the adjusted problem, they would not know that the result is incorrect without actually summing up all the numbers to check, an undesirable chore on a math contest with other problems to worry about.

chongli 7y ago

I'm a pure math student and when I first found the problem I sent it to my friend who is a stats student. She solved it very quickly, but admitted to using a calculator. None of my exams permit calculators, so I've gotten used to never using them. 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.

username90 7y ago
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 the formalism.
- chongli 7y ago
  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 * 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.
  - taesko 7y ago
    I solved it like this: The numbers are x, x+1..., x+10 = 11x + 55. One is erased therefore 10x + 55 - y = 2012. 10x - y = 1957. We know X is whole so we need a number between 1957 and 1967 that divides 10 and that is X. 1960 so X = 196 and therefore y = 3. Answer is X+Y = 199. 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.
  - vworldv 7y ago
    Due to a parity issue (50 here is even) we need to adjust username90's solution. In particular, a sequence of consecutive integers of even length has a half-integer in its middle. 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
    - chongli 7y ago
      Right, but we are talking about sixth-graders here. If a sixth-grader saw username90's solution and tried to apply it to the adjusted problem, they would not know that the result is incorrect without actually summing up all the numbers to check, an undesirable chore on a math contest with other problems to worry about.