Answer:
Step-by-step explanation:
Let us assume that if possible in the group of 101, each had a different number of friends. Then the no of friends 101 persons have 0,1,2,....100 only since number of friends are integers and non negative.
Say P has 100 friends then P has all other persons as friends. In this case, there cannot be any one who has 0 friend. So a contradiction. Hence proved
Part ii: EVen if instead of 101, say n people are there same proof follows as
if different number of friends then they would be 0,1,2...n-1
If one person has n-1 friends then there cannot be any one who does not have any friend.
Thus same proof follows.
I have no clue but using division may be an option
It should take 4 years, please mark me brainliest it would mean a lot and if you want an explanation, Comment and I’ll help you out
4x-5=31
Add 5 to both sides
4x=36
Divide both sides by 4
x=9
Final answer: 9
