You can solve it with no programming. a+b+c = 43 && a^3 + b^3 + c^3 = 17299
As 17299 == 3 mod 4 then as a^3,b^3 and c^3 can be 0 or 1 mod 4 then a, b and c are odd. 27^3 > 17299 so a < b < c <= 25 1^3 == 1 mod 5 2^3 == 3 mod 5 3^3 == 2 mod 5 4^3 == 4 mod 5 5^3 == 5 mod 5 (or 0^3 == 0 mod 5) So as 17299 == 43 + 1 mod 5 at least one of a, b and c must be 2 mod 5 so there are two options, 7 or 17. Case 1 (7): a+b = 36 && a^3 + b^3 = 16956 Case 2 (17) a+b = 26 && a^3 + b^3 = 12386 Let's see Case 1 first: a and b can be 1, 3, 5, 9, 11, 13, 15, 17, 19, 21, 23 or 25, but if a = 1, 3, 5 or 9 then b > 25 so a and b can be 11, 13, 15, 17, 19, 21, 23 or 25. We can easily discard 13 and 23 because of modulo 5 reasons so we have 11, 15, 17, 19, 21 or 25 so we can try with this six numbers. Case 2: a and b can be 1, 3, 5, 9, 11, 13, 15, 17, 19, 21 or 23, but we can discard 3, 13 and 23 again so we have 1, 5, 9, 11, 15, 17, 19 and 21 to try. On Sat, Jul 16, 2011 at 6:54 AM, micke <[email protected]> wrote: > What are the three positive integers whose sum is 43. and the sum of > the cubes of three integers is square of the number 17299. > > -- > You received this message because you are subscribed to the Google Groups > "google-codejam" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/google-code?hl=en. > > -- You received this message because you are subscribed to the Google Groups "google-codejam" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/google-code?hl=en.
