In base 8, twice 12634 is 25470, and twice 25706 is 53614, and thus these serve as alphametic solutions for 2*WRONG = RIGHT. That base has no solutions where 2*RIGHT=WRONG.
This situation seems to be unique to base 8, as the numbers of solutions keeps getting higher for higher bases. Of course it can't be solved in a base lower than 8 as there are 8 different letters used.
In the following table, for each base is shown: the number of solutions of 2*WRONG = RIGHT involving leading zeros; the number that don't involve leading zeros; the number of solutions of 2*RIGHT = WRONG involving leading zeros and those not involving leading zeros.
base
8 0 2 0 0
9 0 3 1 7
10 0 21 5 11
11 0 27 14 33
12 0 49 20 54
13 0 99 31 109
14 0 198 59 164
15 0 211 89 229
16 0 350 117 387
DEFDBL A-Z
CLS
FOR b = 8 TO 16
p4 = b * b * b * b: p3 = b * b * b: p2 = b * b: p1 = b
ztot1 = 0: nztot1 = 0: REDIM used(b - 1)
ztot = 20: nztot2 = 0
FOR w = 0 TO b - 1
used(w) = 1
FOR r = 0 TO b - 1
IF used(r) = 0 THEN
used(r) = 1
FOR o = 0 TO b - 1
IF used(o) = 0 THEN
used(o) = 1
FOR n = 0 TO b - 1
IF used(n) = 0 THEN
used(n) = 1
FOR g = 0 TO b - 1
IF used(g) = 0 THEN
used(g) = 1
FOR i = 0 TO b - 1
IF used(i) = 0 THEN
used(i) = 1
FOR h = 0 TO b - 1
IF used(h) = 0 THEN
used(h) = 1
FOR t = 0 TO b - 1
IF used(t) = 0 THEN
used(t) = 1
wrong = w * p4 + r * p3 + o * p2 + n * p1 + g
right = r * p4 + i * p3 + g * p2 + h * p1 + t
IF 2 * wrong = right THEN
IF w = 0 OR r = 0 THEN ztot1 = ztot1 + 1: ELSE nztot1 = nztot1 + 1
IF b = 8 THEN
PRINT w; r; o; n; g, r; i; g; h; t
END IF
END IF
IF 2 * right = wrong THEN
IF w = 0 OR r = 0 THEN ztot2 = ztot2 + 1: ELSE nztot2 = nztot2 + 1
END IF
used(t) = 0
END IF
NEXT
used(h) = 0
END IF
NEXT
used(i) = 0
END IF
NEXT
used(g) = 0
END IF
NEXT
used(n) = 0
END IF
NEXT
used(o) = 0
END IF
NEXT
used(r) = 0
END IF
NEXT
used(w) = 0
NEXT
PRINT b, ztot1; nztot1, ztot2; nztot2
NEXT
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Posted by Charlie
on 2011-03-04 01:10:35 |
solution