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}.
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