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)"
Basically what we are talking about is the digital root of an integer which is obtained by reducing the number in the mod 9 system and considering its residue. The only exception is the digital root 9 where the mod 9 residue shows up as 0.
Now, we consider the enumerated cubic residues of n in the mod 9 system as follows:
{n, n^3(mod 9) } == {(0,0), (1,1), (2,8), (3,0), (4,1) , (5,8), (6,0), (7,1), (8,8)}
We observe that there are precisely three repeating 3-element cubic residues and these are:
(0,1,8), (1,8,0), and (8,0,1) in the mod 9 system.
Since cubic residue 0 corresponds to 9 as a positive integer digit , we can safely infer that any given 3 consecutive positive integers should satisfy this relationship:
{m^3, (m+1)^2, (m+2)^3 } (- (9,1,8) or, (1,8,9), or, (8,9,1)
Consequently, these three digits when arranged in strictly ascending order of magnitude are 189.
Puzzle Solution
