Subscribe via feed.

Adds up to a Thousand

Posted by Karl Sharman on January 6, 2011 – 3:34 pm

There are 3 different pairs of positive integers that add up to make 6:-

So how many different pairs of positive integers are there to make up 1,000?

As that is too easy, part 2 of the question…. How many different triplets of positive integers add up to make 1,000, and of course, how many different quads of positive integers?

This post is under “Tom” and has 14 respond so far.
If you enjoy this article, make sure you subscribe to my RSS Feed.

14 Responds so far- Add one»

  1. 1. Nathan Said:

    500 unique pairs
    83000 unique triples

  2. 2. Wizard of Oz Said:

    For pairs, 1000/2 = 500

    For each number in the range 1 to 1000 you can make 999 pairs, so 999,000/2 triplets. Each will be triplicated, so divide by 3 to get 166,500 triplets.

    For each of the 500 pairs you can make 998/2 = 499 quads. Each pair will have a duplicate pair in the other set, so the answer is 500*499/2 = 124,750.

    These answers off the top of my head, so probably wrong.

  3. 3. Nathan Said:

    500 pairs
    83,333 triples
    6,965,278 quadruples

  4. 4. cazayoux Said:

    but by what logic and formula do you get the triplets and quads?

  5. 5. Chris Said:

    J cazayoux, if nobody posts the method, I’ll do it tonight. But that’ll be at least 5 hours from now.

  6. 6. Dual Aspect Said:


    I started off this post having worked out that I agreed with Nathan (i.e. triples – 83,333 unique combinations), now I’ve talked myself out of that based on the following……

    If the three numbers making up the total of 1000 were labelled A, B & C then to ensure that every comination is unique A can only have a range of 1-333, just as it has a range of 1-500 for the pairs.

    That leaves us needing to calculate the combinations of B & C, the 2 remaining numbers, for each of the 333 values of A.

    When A=1, B+C = 999 with 2 digits selected from the range 1 to 998, this gives 499 unique combinations.

    When A=2, B+C = 998 with 2 digits selected from the range 2 to 996 (if A=2 then neither B or C can be 1 as this would cause a repeated combination which matches one of those when A=1), this gives 497 unique combinations.

    When A=3, B+C = 997 with 2 digits selected from the range 3 to 994 (as above we cannot use numbers 1 or 2 for values of B or C as this would cause a repeated combination), this gives 496 unique combinations.

    When A=4, B+C = 996 with 2 digits selected from the range 4 to 992, this gives 494 unique combinations.

    When A=5, B+C = 995 with 2 digits selected from the range 5 to 990, this gives 492 unique combinations.

    So a pattern emerges with the number of combinations for B&C reducing by 3 between each subsequent odd / even value of A.

    At that stage my maths knowledge wasn’t enough to express that as a formula so I used a spreadsheet to calculate by brute force. No doubt someone on here can provide a much neater mathematical representation.

    So, after all that, my answer is 83,167.

  7. 7. Dual Aspect Said:

    Amendment – when A=5 there are 493 unique combinations of B and C

  8. 8. Karl Sharman Said:

    Sorry Dual Aspect, I haven’t posted an answer. I do not have a definitive answer, although my calculations agree with Nathan,post 3.

  9. 9. Dual Aspect Said:

    Hi Karl,

    That’s ok, I thought you would agree with Nathan’s answer, I started off in agreement with it too but changed my mind as I was composing the post with my reasoning.

    I’m still uncertain which is right, but my spreadsheet from which I gained my answer seems to provie a neat solution with the number of combinations reducing to just 2 alternatives when A=332 and then just 1 combination when A=333, which is exactly what I would expect it to do.

    So I can’t see the flaw in my calculations, yet I want to believe that it’s 83,333 as it seems like a “nicer” number which sounds more plausible than my 83,167.

    Perhaps Chris can clarify??

  10. 10. Dan Said:

    I realise this was a while ago, but I can confirm that the correct answer is 83,333. My method was:
    Think of 1000 dots in a row. We can partition these into 3 groups by placing 2 ‘barriers’ in the gaps between dots (there will be 999 of these). If you count the number of dots in each section you get 3 numbers that sum to 1000, and all possible choices of ‘barriers’ give all possible ways (999 choose 2 = 498,501).

    But of course this counts every ordering of the numbers as unique, whereas we don’t want to consider the order they are summed in. There are 3!=6 ways to rearrange 3 different numbers so we want to divide by 6. However, some combinations have repeated numbers e.g. 1+1+998 so there are only 3 ways to reorder these (998+1+1 and 1+998+998), so we need to add a ‘duplicate’ for each of these possibilities (i.e. 3 extra for each triple with a repeated number) so when we divide by 6 we get the correct number. There are 499 triples where a duplicate is possible: (1,1,998), (2,2,996),…,(499,499,2). Hence we add 499*3 before dividing by 6 (note that there are no possibilities where all 3 numbers are the same so we do not have to consider the effect of this).

    Hence, number of possibilities is (498,501 + (499*3))/6 = 83,333.

    It took me a while to work out the error in Dual Aspect’s post. In fact, the method is completely sound (to find a formula for the number of combinations instead of brute force, look up ’sum of arithmetic series’: the 167 odd terms are given by 3n-2 for integers n between 1 and 167; their sum is given by (499+1)*167/2).
    The problem lies in a simple counting error for the even values of A; each one has been underestimated by 1. So for A=2 there are 498 possibilities, A=4 has 495 etc. This is because there are as many pairs that sum to 998 as there are to 999 (in general, n and n+1) since you have 499+499=998 and 499+500=999 but not 499.5+499.5.

    Hope this helps.

  11. 11. Meek Said:

    @ Dan what about sum = 27..does your method works??

  12. 12. Yasin Said:


    The code snippet below, count the numbers sequences with non-descending order.
    For example:
    1. {1, 1, 998}
    2. {1, 2, 997}

    499. {1, 499, 500}
    500. {2, 2, 996}

    83333. {333, 333, 334}

    // Written in C#
    int n = 1000;
    int count = 0;
    for (int i = 1; i 0)
    count += temp;


  13. 13. Dennys Said:

    Are you include triplets with repeated numbers like this 994 + 2 + 2 = 1000?

    How many different triplets without repeated numbers whose sum is equal to one thousand are there?

  14. 14. Kaylee Said:

    SO the answer would be 501 because if you do 1 + 999 2 + 998 3 + 997 and so fourth you would do it 500 times but you can’t forget about 0 + 1000 which makes your answer 501. Example: If u were trying to find how many pairs go into 10 you would do, 1+9 2+8 3+7 4+6 5+5 0+10 since there are no more to add your answer would be half of 10 plus 1 which is 6.

Post a reply

PHP Warning: PHP Startup: Unable to load dynamic library 'C:\Program Files (x86)\Parallels\Plesk\Additional\PleskPHP5\ext\php_mssql.dll' - The specified module could not be found. in Unknown on line 0 PHP Warning: PHP Startup: Unable to load dynamic library 'C:\Program Files (x86)\Parallels\Plesk\Additional\PleskPHP5\ext\php_pdo_mssql.dll' - The specified module could not be found. in Unknown on line 0