Yervant1 Posted September 20, 2006 Report Share Posted September 20, 2006 5+3=4^2 Quote Link to comment Share on other sites More sharing options...
Yervant1 Posted September 20, 2006 Report Share Posted September 20, 2006 5+4=3^2 Quote Link to comment Share on other sites More sharing options...
Boghos Posted September 20, 2006 Report Share Posted September 20, 2006 Ok chiefs, give it a crack: the sum of three numbers is 6, the sum of their squares is 8, and the sum of their cubes is 5. What is the sum of their fourth powers? Quote Link to comment Share on other sites More sharing options...
Sip Posted September 21, 2006 Report Share Posted September 21, 2006 (edited) Are you sure about this problem? The closest I get for the sum of fourth powers is something like 20 or 21. But seems like an illposed set of equations with no solution. Edit: There seems to be an exact solution if their sum = 1.7 (sum of squares 8, sum of cubes 5, sum of fourths ~21.6, sum of fifths ~15 ...) Edited September 21, 2006 by Sip Quote Link to comment Share on other sites More sharing options...
karapet Posted September 21, 2006 Report Share Posted September 21, 2006 four plus five equals three to the power of two Quote Link to comment Share on other sites More sharing options...
Azat Posted September 21, 2006 Report Share Posted September 21, 2006 5+3=4^2 wrong Quote Link to comment Share on other sites More sharing options...
Azat Posted September 21, 2006 Report Share Posted September 21, 2006 5+4=3^2 right, but if you were not typing but actually writting it you would not need the ^ Quote Link to comment Share on other sites More sharing options...
Boghos Posted September 21, 2006 Report Share Posted September 21, 2006 Are you sure about this problem? The closest I get for the sum of fourth powers is something like 20 or 21. But seems like an illposed set of equations with no solution. Edit: There seems to be an exact solution if their sum = 1.7 (sum of squares 8, sum of cubes 5, sum of fourths ~21.6, sum of fifths ~15 ...) Sip jan, I am sure ) . Quote Link to comment Share on other sites More sharing options...
Yervant1 Posted September 21, 2006 Report Share Posted September 21, 2006 Boghos Jan I think I can get the values of those three numbers, but it's been a long time and I don't remember all the calculations process. So this is how it goes. Lets call those numbers as a b c We have a+b+c=6 We find the value of a=6-(b+c) insert this in formula two aa+bb+cc=8 [6-(b+c)][6-(b+c)]+bb+cc=8 Find the value of b and insert in formula three. aaa+bbb+ccc=5 In here use the value of a in terms of b and then put the value of b in terms of c once you find the real value of c go bacwards and find b and a . Well I tried Quote Link to comment Share on other sites More sharing options...
Boghos Posted September 21, 2006 Report Share Posted September 21, 2006 Boghos Jan I think I can get the values of those three numbers, but it's been a long time and I don't remember all the calculations process. So this is how it goes. Lets call those numbers as a b c We have a+b+c=6 We find the value of a=6-(b+c) insert this in formula two aa+bb+cc=8 [6-(b+c)][6-(b+c)]+bb+cc=8 Find the value of b and insert in formula three. aaa+bbb+ccc=5 In here use the value of a in terms of b and then put the value of b in terms of c once you find the real value of c go bacwards and find b and a . Well I tried Yervant jan, You did get something right. But not enough. No candy, sorry. Quote Link to comment Share on other sites More sharing options...
Sip Posted September 21, 2006 Report Share Posted September 21, 2006 I see. So the conclusion is that the numbers cannot be just real numbers. This must be some complex "numbers" you want. Quote Link to comment Share on other sites More sharing options...
Yervant1 Posted September 21, 2006 Report Share Posted September 21, 2006 Yervant jan, You did get something right. But not enough. No candy, sorry. For that something right, you can keep the candy just send me a Brasilian Model. Of course "Female" Quote Link to comment Share on other sites More sharing options...
Boghos Posted September 21, 2006 Report Share Posted September 21, 2006 (edited) I see. So the conclusion is that the numbers cannot be just real numbers. This must be some complex "numbers" you want. No Siphan, the conclusion is a natural number, the most modest of them . Added: the numbers are not natural numbers, if that´s what you meant, the sum is. Edited September 22, 2006 by Boghos Quote Link to comment Share on other sites More sharing options...
Boghos Posted September 21, 2006 Report Share Posted September 21, 2006 For that something right, you can keep the candy just send me a Brasilian Model. Of course "Female" Always good to check these days... Quote Link to comment Share on other sites More sharing options...
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