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Date: 11/21/2000 at 16:23:50
From: Stuart
Subject: Recurring decimals

The problem is

     eve/did = .talktalktalk...

Each letter corresponds to a different number, and the fraction 
eve/did is in lowest form.

So far, this is what I have done:

     talk/9999 = .talktalktalk... = eve/did

Dividing 9999 by 3 and 11 gives 303. By trial and error I have found 
that 242/303 gives a four-digit repeating decimal with no digits 
repeated in the fraction or decimal. However, I think this was more 
through luck than judgment. Also, I have to find if there are more 
solutions and, if not, why.

Date: 11/21/2000 at 17:02:13
From: Doctor Rob
Subject: Re: Recurring decimals

Thanks for writing to Ask Dr. Math, Stuart.

Rewrite this in the form

           9999*EVE = DID*TALK
     3*3*11*101*EVE = DID*TALK

Now 101 must be a divisor of the right side. If 101 is a divisor of 
TALK, then you would have T = L and A = K, so that is impossible. Thus 
101 is a divisor of DID. The quotient is D, and this implies that 
D*101 = D0D = DID, so I = 0. This reduces the condition to

     3*3*11*EVE = D*TALK

Now E >= 1, so the left side is at least 121*99 = 11979. This implies 
that D >= 2. But D cannot have a factor in common with EVE, or else 
EVE/DID would not be in lowest terms. Thus D must be a divisor of 99, 
so D = 3 or 9. If D = 9, then 11*EVE = TALK would imply that E = K, 
which is impossible. Thus D = 3, and DID = 303. This reduces the 
equation to

     33*EVE = TALK

This implies that E <= 2. If E = 1, that would make K = 3 = D, an 
impossibility. Thus E = 2. That reduces the equation to

     6666 + 330*V = TALK

so K = 6. Then

     666 + 33*V = TAL

The numbers 0, 2, 3, and 6 are already taken. Thus there are the 
following possibilities for V: 1, 4, 5, 7, 8, or 9. Try each to see if 
you get any duplications of letters.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Puzzles

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