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The only sets of three cubes from among the first 12 cubes that can total to another cube are:
1^3 + 6^3 + 8^3 = 1 + 216 + 512 = 729 = 9^3
3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216 = 6^3
6^3 + 8^3 + 10^3 = 216 + 512 + 1000 = 1728 = 12^3
from
DEFDBL A-Z
DIM cube(12)
FOR i = 1 TO 12
cube(i) = i * i * i
NEXT
CLS
FOR a = 1 TO 12
FOR b = a TO 12
FOR c = b TO 12
t = cube(a) + cube(b) + cube(c)
cr = INT(t ^ (1 / 3) + .5)
IF cr * cr * cr = t THEN
PRINT cube(a); cube(b); cube(c); TAB(30); t; cr
END IF
NEXT
NEXT
NEXT
The first and last contain two duplicates and thus Alice's and Bob's list come from there. The middle one then supplies Carol's list:
27, 64, 125
The totals from the three lists are 729, 2616 and 1728. To add two of these plus a third cube and get another cube, the following program
FOR d = 13 TO 10000
dc = d * d * d
t = 216 + 729 + dc
cr = INT(t ^ (1 / 3) + .5)
IF cr * cr * cr = t THEN
PRINT 216; 729; dc; TAB(30); t; cr
END IF
t = 216 + 1728 + dc
cr = INT(t ^ (1 / 3) + .5)
IF cr * cr * cr = t THEN
PRINT 216; 1728; dc; TAB(30); t; cr
END IF
t = 1728 + 729 + dc
cr = INT(t ^ (1 / 3) + .5)
IF cr * cr * cr = t THEN
PRINT 1728; 729; dc; " "; d; TAB(30); t; cr
END IF
NEXT
finds only
1728 + 729 + 3375 = 5832
where 3375 = 15^3 and 5832 = 18^3.
So Diane's list is 729, 1728,3375.
From Enigma No. 1487, "Fermat plus one", by Adrian Somerfield, New Scientist, 29 March 2008. |
