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Eleven consecutive positive integers are written on a board. Maria erases one of the numbers. If the sum of the remaining numbers is 2012, what number did Maria

Eleven consecutive positive integers are written on a board. Maria erases one of the numbers. If the sum of the remaining numbers is 2012, what number did Maria erase 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.

malaya_zemlya 7y ago
try performing all calculations mod 10.
- chongli 7y ago
  I’m not sure how to make use of your hint. 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).
username90 7y ago
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.
abdullahkhalids 7y ago
The eleven numbers are n,n+1,..., n+10. Their sum is 11n + 55. Let the number removed be n + k, then the sum of the remaining is just 10n + 55 -k = 2012. Or 10n-k = 1957. Then it is obvious that k=3 so that n is an integer. Hence, n=196 The eleven numbers are 196,197,..., 206, out of which 199 was removed.
- chongli 7y ago
  That’s a very good solution. Thanks. 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.
d0mine 7y ago
Given 11 numbers: n, n+1,...,n+10 the sum of 10 of them: 10*n+sum_conseq=2012 where max(sum_conseq)=0+1+...+10=(0+10)+(1+9)+(2+8)..=55 therefore erased number is n+3 because (2012-55-3) divisible by 10 -> n=196.

malaya_zemlya 7y ago

try performing all calculations mod 10.

chongli 7y ago
I’m not sure how to make use of your hint. 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).

username90 7y ago

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.

abdullahkhalids 7y ago

The eleven numbers are n,n+1,..., n+10. Their sum is 11n + 55. Let the number removed be n + k, then the sum of the remaining is just 10n + 55 -k = 2012. Or 10n-k = 1957. Then it is obvious that k=3 so that n is an integer. Hence, n=196 The eleven numbers are 196,197,..., 206, out of which 199 was removed.

chongli 7y ago
That’s a very good solution. Thanks. 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.

d0mine 7y ago

Given 11 numbers: n, n+1,...,n+10 the sum of 10 of them: 10*n+sum_conseq=2012 where max(sum_conseq)=0+1+...+10=(0+10)+(1+9)+(2+8)..=55 therefore erased number is n+3 because (2012-55-3) divisible by 10 -> n=196.