Three neighbors are there. 1st one lends 2nd and 3rd that many no. of tractors that then already each had. After few months, 2nd lends to 1st and 3rd that many tractors then they had. After a few months 3rd lends to 1st and 2nd that many tractors then they had. Now each of them got 24. Find how many they had initially?

Method 1:

Here we are solving problem from the end.

formula:
     Result of after  transaction nth transaction= 1 2 3 4 n
formula for before transaction nth transaction=  1/2 2/2   3/2 4/2 n+1/2+2/2+3/2...

let's apply,
Now each of them got 24

Before 3rd  transaction:
here n is 3.
value for 1st  neighbor =24/2=12
value for 2nd neighbor =24/2=12
value for 3rd neighbor = value of 3rd neighbor+ 1/2 previous value of 1st  neighbor+                                                                             1/2 previous value of 2nd neighbor
                                        =24+24/2+24/2=48
Before 2nd  transaction:
here n is 2.
value for 1st  neighbor =12/2=6
value for 2nd neighbor = value of 2nd  neighbor+ 1/2 previous value of 1st  neighbor+
                                           1/2 previous value of 3rd neighbor
                                        = 12+12/2+48/2=42
value for 3rd neighbor =48/2=24

Before 1st  transaction:
here n is 2.
value for 1st  neighbor = value of 1st   neighbor+ 1/2 previous value of 2nd  neighbor+                                                                           1/2 previous value of 3rd neighbor
                                        =6+42/2+24/2=39
value for 2nd neighbor =42/2=21
value for 3rd neighbor =24/2=12

look at table

Transaction	1st  neighbor	2nd neighbor	3rd neighbor
final	24	24	24
Before 3rd  transaction	24/2=12	24/2=12	24+24/2+24/2=48
Before 2nd  transaction	12/2=6	12+12/2+48/2=42	48/2=24
Before 1st  transaction	12+42/2+24/2=39	42/2=21	24/2=12

so initially 1st had 39 ,2nd had 21 and 3rd had 12 Tractors.

Method 2:

let's assume 1st had x tractors,2nd had y tractors,3rd had z tractors

1st lends y tractors to 2nd and z tractors to 3rd.
So now no of Tractors:
1st: x-(y+z)
2nd: 2y
3rd: 2z

2nd lends x-(y+z) tractors to 1st and 2z tractors to 3rd.
So now no of Tractors:
1st: 2x-2y-2z
2nd: 2y-(x-y+z)
3rd: 4z

3rd lends 2x-2y-2z tractors to 1st and 2y-(x-y+z) tractors to 2nd.
So now no of Tractors:
1st: 4x-4y-4z
2nd: 6y-2x-2z
3rd: 4z -(x+y-3z)

now equating,

4x-4y-4z = 24
x-y-z = 6 ..........(i)

6y-2x-2z = 24
-x+3y-z = 12 .......(ii)

-x-y+7z = 24 .......(iii)

Solving (i) (ii) and (iii) we get,
x = 39, y = 21, z = 12
Hence 39, 21 and 12 tractors each in order for 1st, 2nd and 3rd neighbors