Puzzle #136: Talkative Eve

Another beautiful puzzle from one of the Martin Gardner collections – this one is from “Mathematical Circus” in a chapter titled “Eccentric Chess and Other Problems”.  I am reproducing this verbatim from his collection.

This cryptarithm (or alphametic, as some puzzlists prefer to call them) is an old one of unknown origin, surely one of the best and, I hope, unfamiliar to most readers:

Puzzle #136 Graphic

The same letters stand for the same digits, zero included. The fraction EVE/DID has been reduced to its lowest terms. Its decimal form has a repeating period of four digits. The solution is unique. To solve this, recall that the standard way to obtain the simplest fraction equivalent to a decmal of n repeating digits is to put the repeating digit over n 9’s and reduce the fraction to its lowest terms.

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy Talking!

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4 Responses to Puzzle #136: Talkative Eve

  1. Abhinav Jain says:


  2. Sirji says:

    Hey Alok.. got back on your site after a while.. felt good exercising grey cells
    9999 has the factors 3, 3, 11
    ????/9999 = EVE/DID
    so DID is 101, 303 or 909.
    101 is ruled out because EVE needs to be 3 digit number less than 101
    if DID is 303, EVE can be 2?2 where ? is 1, 4, 5, 6, 7, 8, 9 or 1?1, where ? is 2, 4,5,6,7,8,9
    So we need to try 14 options before assuming DID is 909, which will have 56 options
    By process of elimination, you get to 242/303 = .79867986…
    Not sure if there is a more direct method.

  3. Karan says:

    We’ll proceed after reducing it to the following form.
    Now 9999 has the following prime factors 3,3,11,101 out of which possible 3 digit numbers with similar first and last digits are 101, 303 and 909. 101 can be ruled out as value of the fraction is less than 1 and 101 is the smallest number of the above mentioned form.
    9999/909=11, when EVE*11 the last digit would be E which is not the case, hence 909 can also be rejected. Which leaves us with 303=DID.
    The options for EVE are 1?1, 2?2 (?=[0,9])
    1?1 can be rejected as 1?1*33 will leave K=3 which contradicts D=3.
    We’re now left with 212, 242, 252, 262, 272, 282, 292.
    We now know for sure that K=6 so all we have to check is 212,242,252,272,282,292.
    242/303= .79867986….

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