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The base (or radix) of the number system such that the following equation holds is____________.

$\frac{312}{20} = 13.1$

Let ‘x’ be the base or radix of the number system .

The equation is : $\dfrac{3.x^{2}+1.x^{1} +2.x^{0} }{2.x^{1}+0.x^0} =1.x^{1} +3.x^{0}+1.x^{-1}$

$\qquad \implies \dfrac{3.x^{2}+x +2}{2.x}=x+3 +1/x$

$\qquad \implies \dfrac{3.x^{2}+x +2}{2.x}=\dfrac{x^{2}+3x +1}{x}$

$\qquad \implies 3.x^{2}+x +2=2.x^{2}+6x +2$

$\qquad \implies x^{2}+-5x = 0$

$\qquad \implies x(x-5) = 0$

$\qquad \implies x = 0 \text{ or } x = 5$

As base or radix of a number system cannot be zero, here x = 5.

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base > digit for number system. So x = 5 . It can't be 0(zero) because it violates base > digit constraint.

Ex : (abc)r   here base should be greater than all the digits.

Let x (≠ 0) be the base of the given equation.

We have,

LHS = ( 3x2 + x + 2 ) / ( 2x) = ( 3x / 2 ) + ( 1 / 2 ) + ( 1 / x )

RHS = x + 3 + ( 1 / x )

Now, for the equation to hold true, LHS = RHS

( 3x / 2 ) + ( 1 / 2 ) + ( 1 / x ) = x + 3 + ( 1 / x )

⇒ 3x + 1 = 2x + 6

⇒ x = 5

So, the base is 5.

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can you explain it further and from where we get the equation ??
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Let me explain it like this-

if any number number system is binary and I gave you string 110 then how you will find its value-

Simple 1* 2^2 + 1* 2^1 + 0*2^0

So,we are multiplying by base ^n, n starting from 0 from LSB

Same here we considered base as "x" which we need to find out and formed equaltion,like 20 is there,so it's value is 2*x^1 + 0*x^0.. Like this for all other numbers.

Hope this clarifies your doubt :)
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thankew so much... for nice explanation

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