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) 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?

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 $9999 = 99 \times 101 = 3 \times 3 \times 11 \times 101$ , 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 $A \neq 0$ or 1).

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

$F = E + G$ , $E + F + G \equiv 0$ (mod 3), $A B A \times 33 = E F G$

but $A \neq 0$ , 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