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