If we try to solve the first equation >! $1 / AC = 0.\overline{0BC}~...~(1)$ >! $1000 / AC = BC.\overline{0BC}~...~(2)$ >! >! $(2) - (1) = 999 / AC = BC$ >! $999 = AC \times BC$ For the second equation >! It's the same as the previous one. So >! We need to solve $999 = AC \times BC$. >! As the prime factorization of $999$ is $3^3 \times 37$ and we need both factors ($AC$ and $BC$) to have exactly $2$ digits, hence there is only one possible pair/answer which is $27 \times 37$. Thus >! $A = 2, B = 3, C = 7$ or $A = 3, B = 2, C = 7$