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Math ( Number Theory )

posted by .

How many numbers
from 1 to 1000
inclusive can be
expressed as the sum
of k >= 2 consecutive
positive integers for
some value of k ? Sorry to post question from Brilliant, I got 997 integers, but it's wrong. If you think you can answer ...

  • Math ( Number Theory ) -

    Hmmm.

    For k=2, we want all n<=1000 such that n=m+m+1 = 2m+1
    That is, all odd numbers 3<=n<=999
    There are 996/2+1 = 499 of those

    For k=3, we want n=3m+3
    so, that's all multiples of 3 with 6<=n<=999
    There are 993/3+1 = 332 of those

    For k=4, we want 4m+6
    so, that's all multiples of 4 with 10<=n<=998
    There are 988/4+1 = 248 of those

    There's a pattern here, so you should be able to come up with some


    k=2

    expression, or just brute-force your way through.

    Maybe not a lot of help, but it does show that there are more than 997 of them.

  • Math ( Number Theory ) -

    The ans Steve gave is absolutely wrong...the ans is less than 823 and greater than 665...steve u have counted the same numbers more than one time...for example u have counted 15 twice....

  • Math ( Number Theory ) -

    When I calculated out I got 990 as the correct answer.

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