72 balls were laid out into three unequal heaps of pairs. If we transfer from the first pile to the second as many

72 balls were laid out into three unequal heaps of pairs. If we transfer from the first pile to the second as many balls as there were in this second pile, then transfer from the second to the third as many as in this third pile before, and transfer from the third to the first as many balls as in this first pile then be, then the balls in all piles will be the same number. How many balls were in each pile initially? Print the number of balls in the first, second and third heaps separated by a space.

Let x be the number of balls in the first heap, then y is the number of balls in the second heap and z is the number of balls in the third heap. It is known that there were a total of 72 balls in three piles. Let’s compose and solve a system of equations.
x + y + z = 72
x1 = x-y
y1 = y + y = 2y
y2 = y1-z = 2y-z
z1 = 2z
z2 = z1-x1 = 2z-x + y
x2 = x1 + x1 = 2x-2y
x2 = y2 = z2 = 23
2x-2y = 24
x-y = 12
We output x
x = y + 12
2y-z = 24
Output z
z = 2y-24
2z-x + y = 24
2 (2y-24) – (y + 12) + y = 24
4y-48-y-12 + y = 24
4y = 84
y = 21
x = 21 + 12 = 33
z = 2 * 21-24 = 42-24 = 18
x = 33 y = 21 z = 18
Answer: 33 21 18.