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Pre and Ever Puzzle

  1. Apr 5, 2009 #1
    Solve this cryptarithm, where PRE is a base 10 positive integer and EVER is a base 9 positive integer. Each of the letters represents a different digit, and none of P and E is zero.

    (PRE)10 = (EVER)9 + 1
     
  2. jcsd
  3. Apr 6, 2009 #2
    The smallest that "EVER" can be is 1012, which, in base 9 is 729 + 9 + 2 = 740. That means P >= 7. The biggest that "PRE" can be is 987, which is 1316 in base 9. So E = 1. Further, since "PRE" ends in a 1, it's odd, so "EVER" must be even. In fact, since EVER+1 yields PRE, which is known to end in a 1, R = 0. Also, since 729+4*81 = 1053, V <= 3.

    So, the possible values for EVER are:

    1210
    1310

    Technically, I'm now reduced to brute force, even though there are only 2 possibilities. Is there a deductive way to find the answer at this point?

    DaveE
     
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