Question

1. Show that in any group of 6 people, there is either a set of three mutual strangers or a set of three mutual friends. Is the same statement true for 5 people? Prove or provide a counterexample. 2. Show that, out of any five natural numbers, there are always three the sum of which is divisible by 3.

Can someone solve 8a and 8b without derivatives.

Answer 1

7 let us consider one of them say A. . Among the remaining fire people, suppose that A has at least three friends. If any one two of these three are friends with each other, then together with A, they are a group of three mutual friends. If none of these are friends with each other, then they are a group og three mubual Strangers for themselves." I. If an A has fewer than three friends among the other fire people, then there must be at least three that are strangers to A. If any two of these are strangers, then together with A, they are a group of three strangers. If none of them are Strangers to each other, then they are a group of at least three mutual friends. In case of 5 people, Α . bet straight lines denote friends A & wary lines denote Shuangers. E So, the given figure shows D there are no 3 mutual friends or strangers.

27 Suppose the numbers be X, Rz, dz, Xusky let, Si be the remainders when xy is. divided by 3 for i= 1,2,3,4,5... Now, Ti can be {0,1,2} finc, 0+1+2 = 80 (mod 3) so, all three 10, 1, 2} cannot be in the set g remaindas , R= dvou te , ds. Pats} So, the remainders must be selected from one of the sels 40, 13, 10, 23 or 11,2}. ad by PHP , 7 at least three to numbers whose nemainder will be one of o or I or 2 then, (mod 3) 0+0+0 = 1+1+1 = 2+2+2=0 Hence, the proch.