superprismatic Posted September 3, 2009 Report Share Posted September 3, 2009 cuberoot(3298 + 627550 * sqrt(-5)) + cuberoot(3298 - 627550 * sqrt(-5)) represents a certain positive integer N. If the digits of N are converted to letters (A=1, B=2, etc.), They spell a certain word. What is it? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 3, 2009 Report Share Posted September 3, 2009 (edited) Wonder how I could "aid" in solving this problem! Am I right? Edited September 3, 2009 by DeeGee Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted September 3, 2009 Author Report Share Posted September 3, 2009 Wonder how I could "aid" in solving this problem! Am I right? Right as rain! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 3, 2009 Report Share Posted September 3, 2009 can you hint how you got rid of i? I can't seem to do it ;-) Quote Link to comment Share on other sites More sharing options...
0 bushindo Posted September 3, 2009 Report Share Posted September 3, 2009 (edited) can you hint how you got rid of i? I can't seem to do it ;-) Here's the hint How to take roots of complex numbers- De Moivre's formula Edited September 3, 2009 by bushindo Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 3, 2009 Report Share Posted September 3, 2009 But when i got down to it, it ended up being: 2*cuberoot(3298 +/- i*627550*sqrt(5)) that gives r = 12534*sqrt(12534) and a theta of 89.865 yada yada. which means that the cube root is 97 +/- i*55.9016 or so. Mult by 2, and you get 194 +/- i*111.80 or so. I got as far as that, and that's where my question came in...how did/do you drop the imaginary part so that you could represent all that above as a positive integer??? Quote Link to comment Share on other sites More sharing options...
0 bushindo Posted September 3, 2009 Report Share Posted September 3, 2009 (edited) But when i got down to it, it ended up being: 2*cuberoot(3298 +/- i*627550*sqrt(5)) that gives r = 12534*sqrt(12534) and a theta of 89.865 yada yada. which means that the cube root is 97 +/- i*55.9016 or so. Mult by 2, and you get 194 +/- i*111.80 or so. I got as far as that, and that's where my question came in...how did/do you drop the imaginary part so that you could represent all that above as a positive integer??? I think the bolded part is the problem here I assume you meant that cuberoot(3298 + 627550 * sqrt(-5)) + cuberoot(3298 - 627550 * sqrt(-5)) = 2 * cuberoot(3298 +/- i*627550*sqrt(5)) The above isn't true because you can not distribute outside the cuberoot sign. The straight forward approach would be to compute all 3 roots of cuberoot(3298 + 627550 * sqrt(-5)), then compute all 3 roots of cuberoot(3298 - 627550 * sqrt(-5)). Looking at those, you'll probably find a pair such that their sum is a positive integer with no imaginary part. Computing the 3 cube roots is easier if you draw it on a graph. Edited September 3, 2009 by bushindo Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted September 3, 2009 Author Report Share Posted September 3, 2009 But when i got down to it, it ended up being: 2*cuberoot(3298 +/- i*627550*sqrt(5)) that gives r = 12534*sqrt(12534) and a theta of 89.865 yada yada. which means that the cube root is 97 +/- i*55.9016 or so. Mult by 2, and you get 194 +/- i*111.80 or so. I got as far as that, and that's where my question came in...how did/do you drop the imaginary part so that you could represent all that above as a positive integer??? I'm not sure what 2*cuberoot(3298 +/- i*627550*sqrt(5)) means. But the original problem has it as cuberoot(3298 + i*627550*sqrt(5)) + cuberoot(3298 - i*627550*sqrt(5)). When I did the problem I started by cubing the monster then staring at the result and saw a way to simplify (without De Moiuvre's formula, by the way). I ended up with a cubic equation which I could solve. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 3, 2009 Report Share Posted September 3, 2009 i get (97 + 56i + 97 - 56i) = 194 --> AID Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 3, 2009 Report Share Posted September 3, 2009 I think the bolded part is the problem here I assume you meant that cuberoot(3298 + 627550 * sqrt(-5)) + cuberoot(3298 - 627550 * sqrt(-5)) = 2 * cuberoot(3298 +/- i*627550*sqrt(5)) The above isn't true because you can not distribute outside the cuberoot sign. The straight forward approach would be to compute all 3 roots of cuberoot(3298 + 627550 * sqrt(-5)), then compute all 3 roots of cuberoot(3298 - 627550 * sqrt(-5)). Looking at those, you'll probably find a pair such that their sum is a positive integer with no imaginary part. Computing the 3 cube roots is easier if you draw it on a graph. The equation has 4 answers, 2 of which are the same... cuberoot(3298 + 627550 * i * sqrt(5)) + cuberoot(3298 + 627550 * i * sqrt(5)) cuberoot(3298 + 627550 * i * sqrt(5)) + cuberoot(3298 - 627550 * i * sqrt(5)) cuberoot(3298 - 627550 * i * sqrt(5)) + cuberoot(3298 + 627550 * i * sqrt(5)) cuberoot(3298 - 627550 * i * sqrt(5)) + cuberoot(3298 - 627550 * i * sqrt(5)) The key answers are the middle two...that's where you wind up with principal roots of 97 + 55.901i + 97 - 55.901i = 194 Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted September 3, 2009 Author Report Share Posted September 3, 2009 I cubed N, then found that N3 = 37602 * N + 6596. The number I'm after, then, must be a root of this equation (I may have introduced extraneous roots, though). Solving this, I get 194, -193.824..., and -0.175... I am now done. Voila! Quote Link to comment Share on other sites More sharing options...
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superprismatic
cuberoot(3298 + 627550 * sqrt(-5)) +
cuberoot(3298 - 627550 * sqrt(-5))
represents a certain positive integer
N. If the digits of N are converted
to letters (A=1, B=2, etc.), They
spell a certain word. What is it?
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