List all three-digit numbers such that equal the sum of their first digit and the square of their second digit and the cube of their third digit:
e.g. 135 = 1^1 + 3^2 + 5^3.
This is clearly a simple computer program, or Excel exercise. But I felt like giving it an analytic whirl.
Let the number be abc.
Then 100a + 10b + c = a + b^2 + c^3.
Then 99a + b(10-b) + (c-c^3) = 0
Consider that equation mod 3 and mod 11. This eliminates a as a factor.
I won't bore you with the analysis.
Results are:
Mod 11, bc can only be 00,01,18, 24,29,35,43,63,75,84,89 or 98.
Mod 3, b can never be 2, 5 or 8. That leaves 8 possible values of bc: 00,01,18,35,43,63,75,98.
Calculating b^2 + c^3 determines (for each) the only possible value for a. It is interesting to me that they all lead to solutions, if we allow leading zeroes.
solutions are 000, 001, 518, 135, 043, 063, 175, and 598.
Well, that worked out better than I expected.
Paper and pencil solution (spoiler)
