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 Four or else (Posted on 2010-04-11)
TWO+TWO=NULL
TWO+TWO=ONE
TWO+TWO=FOUR
TWO+TWO=FIVE
TWO+TWO=SIX
TWO+TWO=NINE
Each one of the above mentioned alphametics, treated separately, has more than one base ten solution. Which one has the most? Which one of them, if any, possess an unique solution in another base or bases?

 No Solution Yet Submitted by Ady TZIDON No Rating

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 computer solution | Comment 1 of 9

This table lists the number of solutions for the equation labeled in the header of each column shown, for each base.

`base    NULL ONE FOUR FIVE SIX NINE 5   0     0   0    0    0   0 6   0     1   0    0    0   0 7   0     1   1    0    5   0 8   1     4   2    3    6   1 9   1     9   4   12   25   210   4    12   6   15   42   311   6    18  14   48  101   312   8    21  16   79  154  1013  10    36  19  140  225  1314  14    37  29  197  300  1615  18    53  37  312  476  1316  24    61  45  412  582  2517  28    68  60  568  831  3018  33    82  64  723  977  3519  39   102  78  965 1312  3220  46   104  95 1202 1552  5021  52   133 101 1508 1969  5622  61   145 116 1789 2222  6423  69   160 144 2218 2844  5724  77   173 151 2628 3204  8125  85   209 168 3108 3867  91`

SIX has the most base-10 solutions, 42.

ONE has a unique base-6 solution. ONE and FOUR have unique solutions in base 7. NULL and NINE have unique solutions in base 8, as does NULL in base 9.

DEFDBL A-Z
DIM used(30)
DIM stats(25, 6)
FOR b = 5 TO 25
PRINT b;
FOR t = 1 TO b - 1
used(t) = 1
FOR w = 1 TO b - 1
IF used(w) = 0 THEN
used(w) = 1
FOR o = 1 TO b - 1
IF used(o) = 0 THEN
used(o) = 1
two = t * b * b + w * b + o
two2 = two + two
FOR s1 = 1 TO b - 1
IF used(s1) = 0 THEN
used(s1) = 1
FOR s2 = 1 TO b - 1
IF used(s2) = 0 THEN
used(s2) = 1
one = o * b * b + s1 * b + s2
IF o > 0 THEN
IF two2 = one THEN
stats(b, 2) = stats(b, 2) + 1
END IF
END IF
FOR s3 = 1 TO b - 1
IF used(s3) = 0 AND b > 5 THEN
used(s3) = 1
null = s1 * b * b * b + s2 * b * b + s3 * b + s3
IF s1 > 0 THEN
IF two2 = null THEN
stats(b, 1) = stats(b, 1) + 1
END IF
END IF
four = s1 * b * b * b + o * b * b + s2 * b + s3
IF s1 > 0 THEN
IF two2 = four THEN
stats(b, 3) = stats(b, 3) + 1
END IF
END IF
six = s1 * b * b + s2 * b + s3
IF s1 > 0 THEN
IF two2 = six THEN
stats(b, 5) = stats(b, 5) + 1
END IF
END IF
nine = s1 * b * b * b + s2 * b * b + s1 * b + s3
IF s1 > 0 THEN
IF two2 = nine THEN
stats(b, 6) = stats(b, 6) + 1
END IF
END IF
FOR s4 = 1 TO b - 1
IF used(s4) = 0 AND b > 6 THEN
used(s4) = 1
five = s1 * b * b * b + s2 * b * b + s3 * b + s4
IF s1 > 0 THEN
IF two2 = five THEN
stats(b, 4) = stats(b, 4) + 1
END IF
END IF
used(s4) = 0
END IF
NEXT s4
used(s3) = 0
END IF
NEXT s3
used(s2) = 0
END IF
NEXT s2
used(s1) = 0
END IF
NEXT s1
used(o) = 0
END IF
NEXT o
used(w) = 0
END IF
NEXT w
used(t) = 0
NEXT t
NEXT b

PRINT : PRINT
FOR b = 5 TO 25
PRINT USING "##"; b;
FOR i = 1 TO 6
PRINT USING "####"; stats(b, i);
NEXT
PRINT
NEXT

 Posted by Charlie on 2010-04-11 22:33:28

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