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math plllls heelp

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How many integers 1≤N≤1000 can be written both as the sum of 26 consecutive integers and as the sum of 13 consecutive integers?

  • math plllls heelp - ,

    The sum of 26 consecutive integers starting at n is

    n+(n+1)+...(n+25)
    = 26n + 1+...+25
    = 26n + 25(26)/2
    = 26n + 325

    The sum of 13 consecutive integers starting at m is

    13m + 78

    We want
    26n+325 = 13m+78
    where n <= 25 and m <= 70

    That means that m=2n+19
    Starting with n=1,
    1+...+26 = 351 = 21+...+33
    Now, (1000-351)/26 = 24.9, so our last n will be 1+24 = 25:
    25+...+50 = 975 = 69+...+81

    Looks like there are 25 such integers which are the sum of positive integers.

    If we also allow negative integers, then we can start with n = -12, since

    -12+...+13 = 13 = -5+...+7

    That will add another 13 numbers to the list, making 38 values for N.

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