Find the fraction ABA/CDC =0.DEFGDEFG...

Thank you.Maria Jose.

Well we know that ABA = 101A + 10B, CDC = 101C + 10D and 0.DEFGDEFG = (1000D + 100E + 10F + G)/9999 so:

9999(101A + 10B) = (1000D + 100E + 10F + G) x (101C + 10D)

where A,B,C,D,E,F,G are each one of {0,1,...,9}.

Anyone see where to go from here?

We know 9999=99 ×101 = 3 ×3 ×11 ×101, so we must have D=0 in order for the period 4 decimal expansion. Thus C=3 or 9 (C=1 would give A B A/C D C > 1 for all A ¹ 0 or 1).

If C D C=909, then 1/C D C=11/9999, so A B A×11=E F G. Now, since A=G by considering units digits, so we have a contradiction. Hence C ¹ 9, so C=3.

For C D C=303, we have 1/C D C=33/9999, so, D E F G=E F G is a multiple of 11, but it is also a multiple of 3:

F=E+G, E+F+G º 0(mod 3), A B A×33=E F G

but A ¹ 0, so there are no solutions for which A, B, C, D, E, F, G are all distinct.

Kerwin

Of course, if A,B,C,D,E,F,G are not all distinct, then we could have a lot of possibilities. e.g., A=D=0, C=9 and any choice of B would give an answer in the required form. Also, we do not have to insist on 101 divides CDC, so we could have, say CDC=363 and ABA=242 resulting in the good old ABA/CDC=0.6666....

Kerwin