Excuse me (I used spanish): not in the integers, but apparently there is one in the reals: http://www.wolframalpha.com/input/?i=+a+%2B+b+%2B+c+%3D+43+and+a^3+%2B+b^3+%2B+c^3+%3D+17299^2<http://www.wolframalpha.com/input/?i=+a+%2B+b+%2B+c+%3D+43+and+a%5E3+%2B+b%5E3+%2B+c%5E3+%3D+17299%5E2>
2011/7/16 Alfonso J. Ramos <[email protected]> > No en los enteros, pero al parecer si en los reales: > > http://www.wolframalpha.com/input/?i=+a+%2B+b+%2B+c+%3D+43+and+a^3+%2B+b^3+%2B+c^3+%3D+17299^2<http://www.wolframalpha.com/input/?i=+a+%2B+b+%2B+c+%3D+43+and+a%5E3+%2B+b%5E3+%2B+c%5E3+%3D+17299%5E2> > > > 2011/7/16 Luke Pebody <[email protected]> > >> No answers exist for a + b + c = 43 and a^3 + b^3 + c^3 = 17299^2, >> even allowing them all to be negative: >> http://ideone.com/FTm8l >> >> On Sat, Jul 16, 2011 at 6:28 PM, Leopoldo Taravilse >> <[email protected]> wrote: >> > 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. >> > >> >> -- >> 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.
