If we have the number $(387)_{10} \rightarrow (762)_n$ , how do we calculate the $n$? Thanks in advance.
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HINT:
We have $$387=7n^2+6n+2$$ where integer $n>7$(why?)
lab bhattacharjee
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Well, I do understand that n has to be > 7 because the base 8 is the least it can be since it contains all numbers 0-7. But I'm not sure which one is it exactly.. – user133022 Apr 07 '14 at 14:23
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By solving the equation in bhattacharjee's answer you'll find what $n$ is, do you understand why this works? – Alessandro Codenotti Apr 07 '14 at 14:28
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@user133022, we have $$7n^2+6n-385=0,7n^2+(55-49)n-385=0$$ which has only one integer root $(7)$ – lab bhattacharjee Apr 07 '14 at 14:28
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True, it has 7, but it has to be > 7, it's not >=, therefore it's not in the area of solutions or am I wrong? – user133022 Apr 07 '14 at 14:29
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@user133022, yes, you are right – lab bhattacharjee Apr 07 '14 at 14:32
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I do understand everything you said above, but I still can't find what n is.. any hints for that? Do i have to manually check every single number > 7? Or is there some method to find it easier? – user133022 Apr 07 '14 at 14:34
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@user133022, can't you solve the Quadratic Equation? Else as $762>387$ you can check for $7<n<10$ – lab bhattacharjee Apr 07 '14 at 14:36
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I solved the Quadratic Equation and I got n1=7, n2= ~-8. But it has no meaning to me, right? – user133022 Apr 07 '14 at 14:42
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@user133022, http://en.wikipedia.org/wiki/Negative_base – lab bhattacharjee Apr 07 '14 at 14:43
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Thanks for the link, so I have to take that as base8 right? But does it count as -8 since it's like.. -7,8xxxx not exactly 8? – user133022 Apr 07 '14 at 14:46
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@user133022, Trivially negative or fractional can be discarded – lab bhattacharjee Apr 07 '14 at 14:51
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I'm sorry but I don't understand what am I supposed to do.. English isn't my first language and I have trouble following you.. – user133022 Apr 07 '14 at 14:54
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@user133022, no solution in natural numbers – lab bhattacharjee Apr 07 '14 at 14:55
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So, basically, this has no solution? Is that what you're trying to tell me? The solution is a complex number? But I think this is too complex for me, since we never learned base numbers with complex numbers.. it must be a natural number.. – user133022 Apr 07 '14 at 14:57
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387 in decimal that is 603 in octal and 762 is then base 8 and a fraction
n = 8+k 7n^2 + 6n + 2 = 387 7n^2 + 6n - 385 = 0 is a quadratic with standard form ax2 + bx + c = 0 and when we rearrange x = (-b +- (b^2-4ac)^0.5)/2a substituting x with n n = (-6 +-(6^2-4x7x-385)^0.5)/2x7 n = (-6 +- 104 )/14 that is n= 7 or -7.857142857142857
7x49+6x7+2= 432.14-47.14+2= 387
I now tried also for 0 in decimal and 767 base n see my post https://www.linkedin.com/pulse/proof-0-number-peter-paul-troendle
its really a nice question thanks I had a lot of fun thanks~
