- #151

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Note that any element of the subset ##\{a_{j1}, \cdots, a_{jm}\}## can be represented as a difference of ##s_i## and ##s_{i-1}## using suitable i.

Let ##b_i## be the remainder when ##s_i## is divided by ##n##. ##b_i## can assume n values from ##0## to ##n-1##.

Then $$s_i \equiv b_i~(mod~n)~\cdots~(1)$$

If ##b_i## becomes ##0## then the set ##\{a_1, a_2, \cdots, a_i\}## satisfies the required condition so now we need to consider only those cases where ##b_i## assumes values from ##1## to ##n-1##.

We have ##n## possible sums of the form ##s_i## but only ##n-1## remainders.

So using the Pigeon Hole principle, At least one remainder must have occurred twice.

Say ##b_m = b_n## such that ##m>n##.

Then using ##(1)## we get $$s_m - s_n \equiv 0~ (mod~n)$$

Which means that the required set is ##\{a_{n+1}, \cdots, a_m\}## which will always exist.

And we are done.