• Georgia has three wooden cubes.
• On each face of each cube one of the digits from 0 to 8 is written, with each of those digits appearing on more than one of the cubes.
• The sum of the digits on each cube is the same.
• By moving the cubes around Georgia can make the top faces form many numbers (with the 6 being also used for a 9); including all the three digit perfect squares.
Determine the total number of three-digit perfect cubes that Georgia cannot construct in this way.
Note: Adapted from an Enigma Puzzle which appeared in 'New Scientist' in 1994.
There are 9 digits that each have to appear more than once, and there are 18 faces to be occupied. Therefore, each digit must appear exactly twice. Each cube's total is the same so it must be the twice the sum of the digits 0 through 8, divided by 3. That's 24 total on each cube.
clearvars,clc
potCubes=combinator(9,6,'c')-1;
i=1;
while i<=length(potCubes)
if sum(potCubes(i,:))~=24
potCubes(i,:)=[];
else
i=i+1;
end
end
noPotCubes=length(potCubes);
cubeSets=combinator(noPotCubes,3,'c');
i=1;
while i<=size(cubeSets,1)
tst=potCubes(cubeSets(i,:),:);
for j=1:9
ct(j)=length(find(tst==j-1));
end
if ~isequal(ct,[2,2,2,2,2,2,2,2,2])
cubeSets(i,:)=[];
else
i=i+1;
end
end
for i=10:31
square(i-9)=i^2;
end
square
disp(' ')
for i=1:size(cubeSets,1)
set=cubeSets(i,:);
for j=1:length(set)
disp(potCubes(set(j),:))
end
disp(' ')
end
finds the possible cubes, and lists the 3-digit squares:
The 3-digit squares are:
100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841 900 961
There are two possible sets of three cubes, with the three sets of six numbers each:
0 1 3 5 7 8 Wooden cube 1
0 2 4 5 6 7 Wooden cube 2
1 2 3 4 6 8 Wooden cube 3
0 1 4 5 6 8
0 2 3 4 7 8
1 2 3 5 6 7
All the squares can be formed using the first of the two sets above, but not all can be formed from the second set of three cubes; for example, 400 can't be formed as the 4 would require the use of one of the two cubes having a zero.
The 3-digit perfect cubes are:
125, 216, 343, 512, 729
Any of these five can be formed from the first set of cubes above (or from the second set for that matter, but that's not Georgia's set). The number that she cannot construct is zero.
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Posted by Charlie
on 2023-07-11 14:02:42 |
computer solution
