RE: Brainsteem Mathematics Challenges: Fractions

Thank you, @rycharde. This is interesting. I will try:

1- Based on the assumption ab = (10a+b).
Similarly bc=(10b+c), and so on.

2- Therefore (10a+b)/(10b+a)= (10b+c)/(10c+b)
=> (10a+b) * (10c+b) = (10b+a) * (10b+c)

3- Now we multiply the brackets to get:
(100 a * c ) + (10 a * b) + (10 b * c) + b^2 = 100 b^2 + (10 b * c) + (10 a * b) + (a * c)
=> 100ac + b^2 = 100b^2+ ac
=> 99 ac = 99 b^2
=> a*c = b^2

4- We know that a,b, and c are different single digit numbers. So, we can easily try what number would fit this easy formula: ac = b^2

5- Possible solutions:

a = 1, b = 2, c =4 (because 2^2 = 1*4)
a = 4, b = 2, c =1 (because 2^2 = 4*1)
a = 2, b = 4, c= 8 (because 4^2 = 2*8)
a = 8, b = 4, c= 2 (because 4^2 = 8*2)
a = 9, b = 6, c= 4 (because 6^2 = 9*4)
a = 4, b = 6, c= 9 (because 6^2 = 4*9)