13 Prove by pigeon hole principle that any sequence of 6 who

13. Prove by pigeon hole principle that any sequence of 6 whole numbers has a subsequence whose sum is divisible by 6.

(Hint: consider c₁, c₁ + c₂, c₁ + c₂ + c₃, ... ).

Solution

let us consider 6 whole numbers be

(c1,c1+c2,c1+c2+c3,c1+c2+c3+c4,c1+c2+c3+c4+c5,c1+c2+c3+c4+c5+c6)

which has a subsequence whose sum is divisible by 6

= 6c1 + 5c2 + 4c3 + 3c4+ 2c5 + c6

= 6 ( c1 + 5/6c2 + 4/6 c3 + 1/2 c4+ 1/3 c5 + 1/6 c6)

from here we can easinly have a sum divisible by 6

depending on the values we have taken the sum should be divisible by 6.