Pre and Ever Puzzle

K Sengupta

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

davee123

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