Force distance between 2 charges?

Question

Two charges attract each other with a force of 1.4 N. What will be the force if the distance between them is reduced to one-ninth of its original value?

Answer 1

1.4*9²  = 113.4 N

Answer 2

F1 = K q1q2  / r²

F2 = K q1q2  / (r/9)²

Answer 3

f1=qqk/r^2

so to know the force difference

f2=qqk/(1/9*r)^2

solve for f2

f1/f2 = (1/9 *r)^2/r^2

f1=1.4

1.4/f2=1/81

f2=113.4 N  

Answer 4

F= Kq1q2/r**2

lets say initially r =1m

if r is reduced to 1/9, then r**2 would be 1/81

therefore F=Kq1q2/(1/81) .... F= (81)Kq1q2

the force will be 81 times greater, F= 81(1.4)=113.4 N