I Don’t Mind At All: Mind Your Decisions, on youtube

One of my favourite youtube channels is Mind Your Decisions by Presh Talwalker, a Stanford University graduate.  It’s a collection of math and logic puzzles, presenting solutions to problems using common techniques, or in the case of less common, he explains them in clear language so that the average viewer can understand.  His graphics and presentation are highly polished (even his videos from five years ago), and the mathematics is done on screen, making it easy to follow (or rewind and rewatch if you didn’t get  it the first time).  It’s always easy viewing, and often pleasantly surprising how simple the solutions can be to what appear to be complex problems.  He takes on many “viral” math problems and breaks them down, and shows problems from advanced math competitions and olympiads.

One of his most recent videos was on cube roots and their digits.  How many numbers are there where for any integer x, the sum of the digits of x^3 equals x?  The video preview shows one of them: 17 cubed is 4913, and 4+9+1+3 is 17.  Two of the seven are trivial, 0 and 1.  But what are the others?

Talwalker uses programming to test all the cases, but I’m working on an alternate solution that doesn’t require code and doesn’t involve exhaustive testing, using proof by elimination.  For example, 3√100000 is ~46.42, and 1+9+9+9+9+9 equals 46.  This means there are no solutions with six or more digits because the cube of any other sum will always be larger than the original number.  3√10 is ~2.15, and the cube of 2 is greater than 2, so there are no two digit solutions. (7 and 8 are the only integers whose cube last digit is less than that number, 3 and 2, all others are equal or greater.)  And 3√1000 is 10, so the only three digit solution is 8 (3√512, 5+1+2), since all others produce sums greater than the original number; 3 is less than 7 (3√343) but the sum is greater.  Thus the other four solutions all have four or five digits, but how do you find them without testing? 

Working on it (I know the answers, I’m looking for rules or methods of elimination).  Noticeably, with the except of 10 and 11, in all cases from 2 to 19 the sums of the cubes’ digits are equal to or greater than the original number.  For all cases 21 to 46, all the sums of the cubes’ digits are less than or equal to the original number. All multiples of ten (10, 20, 30, 40) automatically fail, and are less than the solution.