 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Only one number qualifies! (Posted on 2017-11-02) The square of this positive integer equals the sum of cubes of its digits.

Find this number and show that there are no other solutions.

 No Solution Yet Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Answer | Comment 1 of 2

1 is the only solution

2 - 10 are obviously not solutions.

cubes of 1 - 9 are respectively : 1,8,27,64,125,216,343, 512, 729.

For 11 - 19, the sum of digit cubes are 1 + (1,8,27,64,125,216,343, 512, 729) = (2,9,28,65,126,217,344, 513, 730). Other than 9 (but it is not square of 12), they are not even square numbers;

For 21 - 29, the sum of digit cubes are 8 + (1,8,27,64,125,216,343, 512, 729). Only 9 and 16 are square numbers, which are too small to be the square of 21, 22;

For 31 - 39, the sum of digit cubes are 27 + (1,8,27,64,125,216,343, 512, 729). None of them are square numbers.

40+ are not possible, because their squares are always much more than the sum of digit cubes of each

 Posted by chun on 2017-11-02 09:49:16 Please log in:

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