# Ramesh sit around a round table with some other men. He has one rupee more than his right person and this person in turn has 1 rupee more than the person to his right and so on, Ramesh decided to give 1 rupee to his right & he in turn 2 rupees to his right and 3 rupees to his right & so on. This process went on till a person has 'no money' to give to his right. At this time he has 4 times the money to his right person. How many men are there along with Ramesh and what is the money with poorest fellow.

This question is related to Infosys Interview

#### hbt

• May 10th, 2006

I think the problem is stated wrongly.

say, there are 3 persons and let us assume that

a has 5 rupees

b has 4 rupees

and c has 3 rupees  as given in the problem....

But by the problem c should have 1 rupee  more than that of the person sitting at right. ie c should have a+1 rupee

Which never can happen!!!!!!!!!!

correct me if anyone got a better answer.

#### gaurav4067 Profile Answers by gaurav4067

• Dec 7th, 2010

Here, consider there are n number of persons sitting on the table.
Let Ramesh has Rs. x.
Then the person to his right will have (x-1) then to his right will have (x-2)
and so on till the person to the left of ramesh have Rs (x-(n-1))

Here it is said that x gives Re 1 to his right and he in turns give Rs2 to
his right and the process continuess untill we get a person who cannot give to
his right i.e he is let with no money.

In the 1st round person to the right will be left with Rs x-2, to his right
will have x-3 and so on and the person to the left of Ramesh have Rs (x-n)
And now Ramesh has Rs(x +n-1).

In the 2nd round person to the right will be left with Rs x-3, to his right
will have x-4
and so on and the person to the left of Ramesh have Rs (x-(n+1))
And now ramesh has Rs(x + 2(n-1))

Let this process continues for r rounds
Then
person to the right of Ramesh has (x-r-1)
person to the left of Ramesh has (x-(n+r-1))
and Ramesh has (x + r(n-1))

Now we have a condition that we stop when we have a person who has 0 sum
Here person to the left of Ramesh has Rs.0
i.e x=n+r-1

Also ramesh has 4 times the sum of the person to his right
i.e x+r(n-1) = 4(x-r-1)
i.e 3x -4 = 3r + nr
=> 3(n+r-1) - 4 = 3r + nr
=> 3n = nr +7

Now if the process completes in 1 round then r=1
we get 2n = 7
this can not be the number of persons

Now when r=2
we get n=7
=> x = 8

Thus Ramesh has Rs. 8 and the person to his right will have 7,6,5,4,3,2

Proof:
In the 1st Round
Ramesh's right = 6
Ramesh's left = 1
Ramesh = 14

In the 2nd round
Ramesh's right = 5
Ramesh's left = 0
Ramesh = 20

and 20 = 4(5)

#### Suraj Grewal

• Feb 11th, 2012

There are 3 persons sitting with the right hand of Ramesh
assume Ramesh=4
a=3
b=2
c=1
Ramesh gives 1 rs/- to a he has 3 left
now a has 4 and gives 2 to b remaining a has 2
now b has 4 and gives 3 to c remaining b has 1
now c has 4 and gives 4 to Ramesh and remaining c has 0.