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ABA/CDC =0.DEFGDEFG...
By Maria Jose Leon-Sotelo Esteban (T3819) on Wednesday, April 18, 2001 - 02:43 pm:

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

Thank you.Maria Jose.

By Michael Doré (Md285) on Wednesday, April 18, 2001 - 03:19 pm:

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) × (101C + 10D)

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

Anyone see where to go from here?

By Kerwin Hui (Kwkh2) on Wednesday, April 18, 2001 - 04:55 pm:

Are A,B,C,D,E,F,G distinct? If the answer is yes:

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 ABA/CDC>1 for all A¹0 or 1).

If CDC=909, then 1/CDC=11/9999, so ABA×11=EFG. Now, since A=G by considering units digits, so we have a contradiction. Hence C¹9, so C=3.

For CDC=303, we have 1/CDC=33/9999, so, DEFG=EFG is a multiple of 11, but it is also a multiple of 3:

F=E+G, E+F+Gº0(mod 3), ABA×33=EFG

but A¹0, so there are no solutions for which A,B,C,D,E,F,G are all distinct.

Kerwin

By Kerwin Hui (Kwkh2) on Wednesday, April 18, 2001 - 05:01 pm:

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