All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Summing powers (Posted on 2004-07-01) Difficulty: 3 of 5
I thought of three numbers.
Their sum is 6.
The sum of their squares is 8.
The sum of their cubes is 5.
What is the sum of their fourth powers?

See The Solution Submitted by Federico Kereki    
Rating: 3.0000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Imagine | Comment 8 of 18 |

An imaginary number is x + y*i  where i= sqrt(-1)
   The real part is x, the imaginary part is y
That number squared is (x^2 - y^2) + 2*x*y*i
... and  cubed is (x^3 - 3*x*y^2) + (3*x^2*y  - y^3)*i

Start with 3 numbers, each with a Real and Imaginary parts:
a+bi, c+di, e+fi     so there are 6 variables

And we have 3 sums:
6+0i, 8+0i, 5+0i

So we can make 6 equations with 6 unknowns:
1 Sum Realpart N = 6
2 Sum Imaginary N =0
3 Sum Realpart Nsquared = 8
4 Sum Imaginary Nsquared =0
5 Sum Realpart Ncubed =5
6 Sum Imaginary Ncubed =0

1: a+c+e=6
2: b+d+f=0
3: a^2 - b^2 +c^2 - d^2 +e^2 - f^2 =8
4: 2*a*b +2*c*d +2*e*f =0
5: a^3 - 3*a*b^2 + c^3 - 3*c*d^2 + e^3 - 3*e*f^2 =5
6: 3*a^2*b - b^3 + 3*c^2*d  - d^3 + 3*e^2*f  - f^3 =0

So this suggests to me that there is a complex number solution.  Anyone care to solve these 6 equations?


  Posted by Larry on 2004-07-05 02:21:13
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information