Question

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Suppose that T is a tree such that for every vertex v of T. (deg())%3 = 1. Prove that I cannot have 25 vertices.

Answer 1

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Stebl:- T is a tore such that for every Veotex v of T. Cdager)%3=+) Step2 :- degeV) 23 = =1 means a con vertex should be Connect 3k+1' Westex.to get remainder with 1' after modulous of Vertex with (3) Step3 - Degree of any keotex should contain 'I', '4','7', '10' '3k+ 1 to get remainder l' Steby :- 5o, To booke 'T' doesn't contain 25 vertices for 25 vestices formula with be 3*8+1 = 3K+1 Where k=8 of a Steb5 = 25 vertices is in the form of degree vertex sto satisfy (degav) 73=1) So, to Bit Connect 25 Bestices, we require Tetra kestices whose deg will be as and remainder will be all after mod 3.

Stepo - Example: 25 20 -deg(25) 30 Total 1261 votes vestices tequired Step7 - Conclusion "We tried to get 25 vertices in a tree but it's not bossible. Total. vertices will be more than 25 08 thach 고 less, 25 but not exactly 25 because fails to meet condition we Step 8 i To meet the condition (deger mod 3=1 Total kertices should be in the se foom of 3k+2 = 3ktl for degree and extoa 4 for connecting those kertices Hence, we can clearly see that 25 vertices is not multiple of (3K+2'. Hence, proved