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$$213_x=139_{10}$$

$$21_x=1021_4$$

How would I solve this

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What do quadratic forms have to do with this? –  Marc van Leeuwen Mar 11 '13 at 15:16
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2 Answers 2

I'm not sure wether i should reply or not since this smells like homework; anyway, just think what base x means:

$213_x = 139_{10} \Rightarrow 2x^2+x+3 = 139$

$21_x = 1021_{4} \Rightarrow 2x + 1 = 4^3 + 2\cdot 4 + 1$

Now solve the 2 equations and you're done

PS : on the right side of the $\Rightarrow$ , every number is assumed to be $\text{base 10}$

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$$213_x=139_{10}$$ $$2\cdot x^2+1\cdot x^1+3\cdot x^0=1\cdot 10^2+3\cdot 10^1+9\cdot10^0$$ $$2x^2+x+3=139$$ $$2x^2+x-136=0\Rightarrow x_{1,2}=\frac{-1\pm\sqrt{1+8\cdot 136}}{4}=\frac{-1\pm33}{4}$$ $$x_1=8$$ the base $$x_2=-17/2$$ is not a correct base see comments below.

similarly $$21_x=1021_4$$ $$2x+1=4^3+2\cdot 4 +1=73\Rightarrow x=36$$

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Bases are normally positive integers only, sometimes negative integers make sense. Rationals I've never seen. –  vonbrand Mar 11 '13 at 18:03
    
you never seen, me too but that does not mean that they do not exist, the example above shows that they exist. Can you give the reason that such bases are wrong to use. –  Adi Dani Mar 11 '13 at 18:13
    
They are wrong because $-1111_{-2} = 101_{-2}$ –  kaharas Mar 11 '13 at 18:22
    
@AdiDani: Nice Adi, have a good day ahead. –  Babak S. Mar 14 '13 at 6:12

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