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You don't need algebra. You know that 10 numbers are 2012, so on average 201, so the middle of the numbers needs to be around there. After that you just try a f

You don't need algebra. You know that 10 numbers are 2012, so on average 201, so the middle of the numbers needs to be around there. After that you just try a few sequences around 201 and you'll find it pretty quickly, most can add 11 numbers pretty fast in 6'th grade. Forcing yourself to only rely on the solution strategies taught by school makes you very bad at problem solving.

chongli 7y ago
I specifically set out to find a solution that doesn’t involve any brute force (trial and error). Trial and error can waste all of your time and get you nowhere on nontrivial problems.
- mepiethree 7y ago
  OTOH, puzzling something out through trial and error fosters curiosity and teaches a "fail fast" mindset. For a sixth grader, this "play" may be a more beneficial lesson than only being able to tackle problems for which you already know the optimal strategy (in this case, algebra)
  - chongli 7y ago
    I disagree. The hard problems in math, science, and engineering are not amenable to a "fail fast" mindset. They take years, decades, or sometimes centuries of study to solve. This sort of puzzling through play is good for solving math contest problems that have been specifically designed for the purpose, and little else.
    - d0mine 7y ago
      Simple examples may be useful. Exploring them, proving stuff about them it is how one may carve a foothold to solve a harder problem. Example: The Josephus Problem - Numberphile https://youtube.com/watch?v=uCsD3ZGzMgE&t=86 https://youtube.com/watch?v=uCsD3ZGzMgE&t=86
- username90 7y ago
  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.
- SamReidHughes 7y ago
  The act of figuring out how to solve the problem is itself an act of trial and error. And in many problems, the clever part is figuring out how to target your trial and error. 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.

chongli 7y ago

I specifically set out to find a solution that doesn’t involve any brute force (trial and error). Trial and error can waste all of your time and get you nowhere on nontrivial problems.

mepiethree 7y ago
OTOH, puzzling something out through trial and error fosters curiosity and teaches a "fail fast" mindset. For a sixth grader, this "play" may be a more beneficial lesson than only being able to tackle problems for which you already know the optimal strategy (in this case, algebra)
- chongli 7y ago
  I disagree. The hard problems in math, science, and engineering are not amenable to a "fail fast" mindset. They take years, decades, or sometimes centuries of study to solve. This sort of puzzling through play is good for solving math contest problems that have been specifically designed for the purpose, and little else.
  - d0mine 7y ago
    Simple examples may be useful. Exploring them, proving stuff about them it is how one may carve a foothold to solve a harder problem. Example: The Josephus Problem - Numberphile https://youtube.com/watch?v=uCsD3ZGzMgE&t=86 https://youtube.com/watch?v=uCsD3ZGzMgE&t=86
username90 7y ago
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.
SamReidHughes 7y ago
The act of figuring out how to solve the problem is itself an act of trial and error. And in many problems, the clever part is figuring out how to target your trial and error. 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.