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Only one number qualifies! (Posted on 2017-11-02) Difficulty: 3 of 5
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    
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proof Comment 2 of 2 |
N = number sought for.  It has d digits.

The largest N contains all 9s and the smallest equals 10^(d-1).

So 10^(d-1) <= N < sqrt((9^3)*d) = 27*sqrt(d)

A 3-digit number is impossible since 100 > 27*sqrt(3) < 47 and therefore bigger numbers are impossible as well.

For d=2, N < 38.  But 4^3 > 38 so we only have to check 30-33. Similarly, 20-22 and 10-12 are quickly eliminated.

For d=1 only N=1 works and it is the sole solution.

  Posted by xdog on 2017-11-02 17:55:37
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