SOLUTION: I have a three digit number such that the sum of the cubes of each digit gives the three digit number containing the same digits. Find my number. I did, x^3 + y^3 + Z^3 = 100x

Question 876581: I have a three digit number such that the sum of the cubes of each digit gives the three digit number containing the same digits. Find my number.
I did, x^3 + y^3 + Z^3 = 100x + 10y + z
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
There are several:

153 (1^3 + 5^3 + 3^3 = 153)
370
371
407

The equation x^3 + y^3 + z^3 = 100x + 10y + z where x,y,z are digits is correct, but I don't think there's an easy algebraic solution (particularly when the correct numbers are 153, 370, 371, 407). It is far easier to write a computer program to solve it, since there are only 900 cases to check. The following Python code solves it (without _'s)
for i in range(1,10):
__for j in range(0,10):
____for k in range(0,10):
______n1 = i**3 + j**3 + k**3
______n2 = 100*i + 10*j + k
______if n1 == n2: print n2
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