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Getting Cubic With T^2 (Posted on 2007-10-10) Difficulty: 3 of 5
Determine all possible positive integers T such that T2 is equal to the cube of the sum of the digits of T.

See The Solution Submitted by K Sengupta    
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Solution computer analysis/solution | Comment 2 of 4 |

We have two functions of t: t^2 and the cube of the sum of the digits of t. We want to know for what values of t these two functions are equal.  If we seek the answers by exhaustively looking at values of t up to a certain point, we want to be assured that no more solutions will lie above that value of t.

If we let n = the number of digits in t, and specify these functions, expressed in terms of n, for the upper limit of either the ratio or the difference between the cube of the s.o.d. and the square of the original number, that upper limit is reached when all the digits are 9, as in 9, 99, 999, etc.

The square of t is going to be very close to 10^(2*n), while the cube of the sum of the digits is at most 729*n^3.  The exponential increases, of course, exponentially with n, while the cubic is merely that, a cubic in n.  Here's a table of values of t^2 vs. the cube of the sum of the digits, at those places (all-9's) where the cubic has the best chance against the exponential in n:

    t             t^2        cube of sum of digits 
        9                 81          729
       99               9801         5832
      999             998001        19683
     9999           99980001        46656
    99999         9999800001        91125
   999999       999998000001       157464
  9999999     99999980000001       250047
 99999999   9999999800000001       373248
999999999 999999998000000001       531441

If we test t up to 1,000,000, we've passed the point in the table where we got

   999999       999998000001       157464

so we can safely say we've got them all, as the smallest t^2 for a 7-digit number is 10^12, while the largest cube of s.o.d. for a 7-digit number is 250,047.

In fact, we could have stopped with 3-digit numbers, as the smallest t^2 for a 4-digit number is 1,000,000, while the largest cube of s.o.d. for a 4-digit number is 46,656.

FOR i = 1 TO 1000000
  sq = i * i
  n$ = LTRIM$(STR$(i))
  t = 0
  FOR j = 1 TO LEN(n$)
   t = t + VAL(MID$(n$, j, 1))
  cube = t * t * t
  IF sq = cube THEN PRINT i, sq, cube

which gives two solutions: 1 and 27:

  1             1             1
 27            729           729

Edited on October 10, 2007, 11:00 am
  Posted by Charlie on 2007-10-10 10:44:56

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