For every cube of an integer, the recursive sum of its digits , e.g. 729 -> 18 -> 9 etc. is always 1,8 or 9. With a computer program i checked this phenomenon up to 1000000000. In my prior question(link-> A mathematical phenomenon regarding perfect squares….) i asked the mathematical reason behind similar phenomenon regarding perfect squares. Now i wonder what will be the reason behind this cube phenomenon? Any answer?
Some of the last outputs of cubes and their recursive sum of my program are below:
970299000 Recursive sum is 9
973242271 Recursive sum is 1
976191488 Recursive sum is 8
979146657 Recursive sum is 9
982107784 Recursive sum is 1
985074875 Recursive sum is 8
988047936 Recursive sum is 9
991026973 Recursive sum is 1
994011992 Recursive sum is 8
997002999 Recursive sum is 9
If we go further…. For every 4th power of an integer the recursive sum of digits is either 1,7,9 or 4.
For every 5th power of an integer the recursive sum of digits is either 1,2,4,5,7,8 or 9.
Can we derive some formula that can tell the recursive sum of digits of n'th power of an integer?
Best Answer
A positive integer is either $3k$ or $3k\pm 1$ for some integer $k$.
$(3k)^3=27k^3=9(3k^3)$ is a multiple of $9$.
$(3k\pm 1)^3=27k^3\pm 27k^2+9k\pm 1=9(3k^3\pm 3k^2+k)\pm 1$ is of the form $9l\pm 1$ for some integer $l$.
Together with the fact that a positive integral multiple of $9$ has recursive sum $9$, this proves your conjecture.