Eve Did Talk Cryptarithm

Looking up cryptarithms – or puzzles where letters substitute for numbers and you have to solve the puzzle by mapping the letters to digits \(0-9\) so that the actual math comes out right – for puzzle club, I came across this one, called the Eve Did Talk cryptarithm or alphametic:


After initially trying to solve it by asking Perplexity (an AI chatbot) to write a Javascript script for it, I determined that this route would be too troublesome. Instead, I decided to try and tackle it by eliminating most possibilities, and seeing what happened. I started by looking at a reference for restating repeating decimals as fractions, which is what this alphametic is actually looking for. It turns out that any repeating decimal can be restated as a fraction by using the repeating digits in the numerator of a fraction and as many \(9\)s as necessary to equal the amount of repeating digits. For this example, we have, for the right side of the equation:

\[.\overline{TALK} = \frac{TALK}{9999}\]

where T, A, L and K stand for unique digits between \(0\) and \(9\).

The next step for me was to determine all the possible values of the LHS denominator by dividing \(9999\) by an integer such that a 3-digit integer resulted. I could only find 3 integers that divided \(9999\) evenly:

\[9999/11 = 909\]

\[9999/33 = 303\]

\[9999/99 = 101\]

The \(101\) in the LHS denominator I eliminated right away, since the only possible numerator would be \(100\), and that wouldn't satisfy the cryptarithm constraints, and I also determined that the \(909\) as LHS denominator wouldn't work in the next step.

The next step was to set up the equation with the denominators we determined previously and numerators that were unknown, but related to each other (I've eliminated the \(101\) denominator case):

\[\frac{x}{303} = \frac{33x}{9999}\]

\[\frac{x}{909} = \frac{11x}{9999}\]

Next I tried to find some possible LHS numerators that satisfied the constraints given in the problem statement, which are that the first and third digits are the same and not equal to any of the digits of the LHS denominator. I quickly figured out that the LHS denominator case of \(909\), where you are multiplying the LHS numerator by \(11\), will always give the LHS ones digit in the RHS numerator's last digit, and that wouldn't work, because the RHS numerator didn't have any digits the same as the LHS numerator or denominator. So then I worked on the RHS denominator case of \(303\), and tried \(252\) for the LHS numerator, which didn't work, and then \(242\), which did! Fortunately, there aren't many possibilities for the LHS numerator, since the number had to be less than \(303\) (since the fraction had to be less than \(1\)) and had to have the first and last digit the same, and not \(0\) or \(3\).

When you substitute in \(242\) for \(x\), you get the following solution:

\[\frac{242}{303} = \frac{7986}{9999}\]

So one solution to this alphabetic is: \[\boxed{E = 2, V = 4, D = 3, I = 0, T = 7, A = 9, L = 8, K = 6}\].

Can you find other solutions?

Check out this great paper on cryptarithms and alphametics for more puzzles and solutions.