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Maximum and cubic to a T (Posted on 2008-10-09) Difficulty: 3 of 5
Find analytically the maximum value of a positive base 10 integer T, such that T is equal to the cube of the sum of its digits.

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

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Solution solution Comment 6 of 6 |
For a cube root of 1 digit, the maximum number of digits of the cube is 3 digits.
For a cube root of 2 digits, the maximum number of digits of the cube is 6 digits.
For a cube root of 3 digits, the the maximum number of digits of the cube is 9 digits.

The maximum number of digits of a sum of a 9-or-less-digit number is 2 digits.

The rate at which the maximum number of digits of a sum of the digits of a number increases is much less than the rate at which the maximum number of digits of a cube increases.
Therefore, the maximum number is where the maximum number of digits is 6 digits, where the maximums are equal, or less.

As the maximum sum of digits of 6-digit cubes (cube roots of 47 to 99) is 36 or less, and as 36 is less than 47, the maximum number of digits must then be a 5-digit number or less.

The sums of the digits of 5-digit cubes (cube roots of 22 to 46) are {9, 10, 17, 18, 19, 26, 27, 28}, therefore, for a 5-digit solution, only cubes with sum totals of 26, 27 and 28 are possible.

283 = 21952. The sum of the digits 2+1+9+5+2 is only 19, thus, the cube root must be less. 

273 = 19683. The sum of the digits 1+9+6+8+3 is 27, which equals the cube root, therefore a 5-digit cube is a solution and the 5-digit cube is 19683.
  Posted by Dej Mar on 2008-10-10 06:34:33
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