What 2-digit number(s), when squared, produces a number equal to the cube of its sum of digits?
27 is the only such number.
My method:
Call the digits a and b so the square is (10a+b)^2 and the cube is (a+b)^3
(10a+b)^2 = (a+b)^3
100a^2 + 20ab + b^2 = a^3 + 3a^2b + 3ab^2 + b^3
a^3 + (3b-100)a^2 + (3b^2 - 20b)a + (b^3 - b^2) = 0
Which is a cubic in a. While this is solvable in principle, in practice it was easier to make an array in Excel and see where the sum was zero.
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Posted by Jer
on 2012-11-29 11:27:27 |