How to find the radix in number system.?

Question

I have $(132)_{10} = (2010)_r$

I have tried the above .. I got the answer that $2(r^3)+r =132 $

From this I am unable to find the value of $r$. Can anyone help me out to solve this problem?

Answer 1

Hint: $f(r) = 2r^3+r$ is an injective function, and $(2010)_{10} > (132)_{10}$. That means you only have to check $r=1$, $r=2$, etc. and see which one matches. Binary search is faster, and it's obvious that $r$ must be even (because $f(r)$ is odd for odd $r$).

Answer 2

Note that $2r^3+r$ is an increasing function.

By doing binary search, we can see that

$$2(4^3)+4=2(64)+4=132$$

Answer 3

Note that $$\sqrt[3]{\frac12\cdot(2010)_r}\approx \sqrt[3]{\frac12\cdot(2000)_r}=\sqrt[3]{(1000)_r}=(10)_r=r$$

Answer 4

Our goal is to find a positive integer solution to $2 r^3 + r - 132 = 0$.

To proceed, we use the rational root theorem, which states that any rational solutions $\frac{p}{q}$ must have $p$ dividing $-132$ and $q$ dividing $2$.

Factoring $132$ gives you $2^2 \cdot 3 \cdot 11$. So here are all the possibilities (ignoring sign):

$q = 1$ or $2$

$p = 2$ or $3$ or $4$ or $6$ or $11$ or $\dots$ or $132$

Now observe $4$ is a solution to $2 r^3 + r - 132 = 0$.