Find all positive integers n < 1000 such that the cube of the sum of the digits of n equals n2.
Solution
n < 1000, so the sum of the digits is at most 27, so n2 is a cube not exceeding 273. So we are looking for m3 which is also a square. That implies m is a square. So the only possibilities are m = 1, 4, 9, 16, 25. Giving n = 1, 8, 27, 64, 125. The corresponding cubes of the digit sums are 1, 512, 729, 1000, 512, whereas the corresponding squares are 1, 64, 729, 4096, 15625. Thus the only solutions are n = 1, 27.
© John Scholes
jscholes@kalva.demon.co.uk
5 Aug 2002