Consider positive integers which are perfect cubes, do not end in zero, are not palindromes, and their reverse is also a perfect cube.
Q1: Find the smallest such reversible cube which also contains a string of 4 consecutive zeros.
Q2: What do the following pairs of integers have to do with these reversible cubes, and what pair would come next? (hint: see Conjecture 1)
(11, 13), (19, 25), (35, 49), (37, 41), (39, 57), (43, 53), (67, 97)
Conjectures:
(1) the cube roots of all reversible cubes contain only two unique digits, 'a' and 'b'.
(2) the first digit of all reversible cubes is 'c' and the next non-zero digit is always 'd'.
Bonus Questions related to Conjectures:
What are a and b; and what are c and d?
Can you prove the conjectures to be either true or false? (I have not been able to do so)
If a proof is not possible, can you offer an explanation for this finding? (I cannot)
clearvars, clc
for n=1:90000
cube=n*n*n;
c=char(string(cube));
if c(end)~='0'
c2=flip(c);
cube2=str2num(c2);
if cube2~=cube
cr2= round(cube2^(1/(3)));
if cr2*cr2*cr2==cube2
disp([n cube cr2 cube2])
end
end
end
end
finds these reversible cubes, each preceded by its cube root.
1011 1033364331 1101 1334633301
1101 1334633301 1011 1033364331
10011 1003303631331 11001 1331363033001
11001 1331363033001 10011 1003303631331
So, 1st Bonus Question first:
It looks like the cube roots contain only zeros and 1's. These are the a and b of the puzzle in some order.
The c and d are respectively 1 and 3.
Time constraints prevent using the slower vpa (variable precision arithmetic) to continue with the above logic, but using another conjecture, that the cube roots will be in the form 1100...001 and 100...0011, where the zeros are collapsible down to one of them and expandable indefinitely and that there are no other arrangements that work, another program:
clearvars
digits 300
for z=1:9
crs=['1' repmat('0',1,z) '11'];
cr1=vpa(crs);
cube1=cr1^3;
crs=flip(crs);
cr2=vpa(crs);
cube2=cr2^3;
disp([cr1 cube1 cr2 cube2])
end
finds
1011, 1033364331, 1101, 1334633301
10011, 1003303631331, 11001, 1331363033001
100011, 1000330036301331, 110001, 1331036300330001
1000011, 1000033000363001331, 1100001, 1331003630003300001
10000011, 1000003300003630001331, 11000001, 1331000363000033000001
100000011, 1000000330000036300001331, 110000001, 1331000036300000330000001
1000000011, 1000000033000000363000001331, 1100000001, 1331000003630000003300000001
10000000011, 1000000003300000003630000001331, 11000000001, 1331000000363000000033000000001
100000000011, 1000000000330000000036300000001331, 110000000001, 1331000000036300000000330000000001
(matrix brackets and .0 removed to condense the output)
So my own conjecture as to the layout of the zeros and 1's would give an answer to Q1: 1000033000363001331, which of course is smaller than its reverse 1331003630003300001.
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Posted by Charlie
on 2021-06-29 11:34:06 |
computer exploration
