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)
The first pair of reversible cubes I found were:
1011^3=1033364331
1101^3=1334633301
The next
10011^3=1003303631331
11001^3=1331363033001
(I only searched to 23938, but the above extends to)
100011^3=1000330036301331
110001^3=1331036300330001
1000011^3=1000033000363001331
1100001^3=1331003630003300001
Which I presume to be the solution to Q1.
It appears the conjectures are that the cube roots contain only 0 and 1, whereas the cubes start with 1 and the next non-zero digit is 3.
Google Sheet I used:
https://docs.google.com/spreadsheets/d/1qrAgMxmFGLTN722ZCPSS3l3-PuIN7TXAZhlLruqgKMY/edit?usp=sharing
|
Posted by Jer
on 2021-06-29 10:24:55 |