Question

Ten girls, numbered from
1 to 10 , sit at a round table, in a random order. Each girl then
receives a new number, namely the sum of her own number and those of her two neighbours.
Prove that some girl receives a new number greater than 17.

Answer 1

are the no.s integers?

Answer 2

ya real numbers

Answer 3

@vikrantg4 My friend help me

Answer 4

Will try my best bro @goformit100

Answer 5

My genius friend @samigupta8 Help me please .

Answer 6

I can't even figure out the basis to solve it :/

Answer 7

No Worries, you can proceed with your work.

Answer 8

Have you started with something so far ?

Answer 9

no I am unable to figure it out too

Answer 10

Divide the nine girls with numbers 2 to 10 into three groups of
three: the three girls to the left of no 1, the three girls to the right of no 1, and the three girls
opposite no 1. The total sum of their numbers is 54, so the sum of at least one group must
be at least 18. Therefore the new number of the girl in the middle of that group is at least 18.

Answer 11

ok then ?

Answer 12

i just gave this answer.

Answer 13

Thank you Sir

Answer 14

Greater than 17? or greater than or equal to 17?

Answer 15

If the latter, it's trivial... Acknowledge that the sum of our new numbers must be \(3\) times our original sum (this is only intuitive), which is \(3\times(1+2+3+\dots+10)=3\times55=165\). It's clear that our average new assignment number would be \(165/10=16.5\) It's clear then that some person must have at least \(17\).

Answer 16

Consider knowing some person has \(1\); consider the sums around the table of groups of 3, which yield the total sum \(2+3+4+\dots+10=54\). The average sum of these groups is then \(54/3=18\), so one of our people ends up assigned with at least \(18\).