Solution to Problem 4: "Digitalis"

(The paragraph marked [CJR] was suggested by Carl Johan Ragnarsson, as it was much clearer than the means I had used to find D.)

There are six equations to satisfy:

(I)    IFIB - CCE = EFF (or EEF + CCE = IFIB)

(II)    EBG - GD = EED

(III)    CEH + FE = CBA

(IV)    IFIB - EBG = CEH

(V)    CCE / GD = FE  (or FE * GD = CCE)

(VI)    EFF + EED = CBA

From (I), we see that I = 1. 

[CJR] Examining the units digit of (V), we see that E*D = E (modulo 10).  There are four ways that this could happen: E = 0, E = 5, D = 1, and D = 6.  E cannot be 0, because it is the first digit of EFF.  E cannot be 5, because the sum in (VI) would have to be a four-digit number.  D is not 1, because we know that I is 1.   Therefore, D = 6. 

Looking at the unit's digit of (II), G = 2.  If F were 4 or greater, then the product in (V) would be greater than three digits, so F = 3.  From the unit's digit of (III), A = 9.  CCE = 26 * 3E is at least 800, and since 9 is taken, C = 8.   By (III), E = 4 and B = 7.  Finally, by (VI) (or process of elimination), H = 5.

The solution, giving the values of A-I in order is 9 7 8 6 4 3 2 5 1.

Comments: I barely looked at this puzzle in the contest.  It is like the Cross Sums puzzle in that it is something that I've been doing in Dell puzzle magazines for decades, but couldn't work it out under time pressures.