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Getting 189 (Posted on 2016-09-05) Difficulty: 1 of 5
Evaluate cubes of three consecutive positive integers.
Add up the digits in each of the three results, and add again until you’ve reached a single digit for each of the three numbers.
For example:
46^3 = 97336; 9 + 7 + 3 + 3 + 6 = 28; 2 + 8 = 10; 1 + 0 = 1
47^3 = 103823; 1 + 0 + 3 + 8 + 2 + 3 = 17; 1 + 7 = 8
48^3 = 110592; 1 + 1 + 0 + 5 + 9 + 2 = 18; 1 + 8 = 9

Putting the three digits in ascending order will always give the result 189.
Please explain why.

Source:Kendall and Thomas "Mathematical Puzzles for the Connoisseur (1971)"

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts Puzzle Thoughts Comment 3 of 3 |
(In reply to Puzzle Solution by K Sengupta)

We know that in mod 7 system, for any given integer n, we must have:

n^3 (- {0,1,6} (mod 7)
Accordingly,  the cubic residue cycle {0,1,1,6,1,6,6} repeats indefinitely.  However, no consecutive 3 elements of the foregoing cycle repeats indefinitely and accordingly,  we cannot utilize the above result for our desired proof.

  Posted by K Sengupta on 2022-06-10 23:38:32
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