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Four or else (Posted on 2010-04-11) Difficulty: 3 of 5
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    
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Solution 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   2
10   4    12   6   15   42   3
11   6    18  14   48  101   3
12   8    21  16   79  154  10
13  10    36  19  140  225  13
14  14    37  29  197  300  16
15  18    53  37  312  476  13
16  24    61  45  412  582  25
17  28    68  60  568  831  30
18  33    82  64  723  977  35
19  39   102  78  965 1312  32
20  46   104  95 1202 1552  50
21  52   133 101 1508 1969  56
22  61   145 116 1789 2222  64
23  69   160 144 2218 2844  57
24  77   173 151 2628 3204  81
25  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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