Divisibility

Tagged: 

Viewing 2 posts - 1 through 2 (of 2 total)
  • Author
    Posts
  • #66010
    Shreya Nair
    Participant

    Divisibility

    Attachments:
    You must be logged in to view attached files.
    #66788
    Shirsendu Roy
    Moderator

    a1,a1+a2,…..,a1+a2+…a100

    if any one of these 100 terms divisible by 100 then we are through.

    otherwise suppose that none of them is divisible by 100, then each leaves a remainder 1,2,…99. Since there are 100 sums and 99 remainder, by Pegion hole principle PP1, two of the sums leaves the same remainder after division by 100,

    then let a1+a2+….+am=b100+r

    a1+a2+….+an=c100+r

    That gives m<n, a_{m+1}+a_{m+2}+….+a_{n}=100(c-b)

    or, 100 divides a_{m+1}+a_{m+2}+….+a_{n}.

Viewing 2 posts - 1 through 2 (of 2 total)
  • You must be logged in to reply to this topic.