How to find the base r?

Question

$\sqrt{144_r} = 12_r$

What is r?

The method I used is:

$\sqrt{ ((1 × r^2) + (4 × r^1) + (4 × r^0))} = ((1 × r^1) + (2 × r^0))$ and I tried solving this equation but I got now where.

the solution to this question according to the book is $r\geq 5$.

Answer 1

Your attempt is a good start - and actually you are almost there!

As you have written, we manipulate the LHS: $$\begin{align*} \sqrt{(144)_r}&=\sqrt{r^2+4r+4} \\ &= \sqrt{(r+2)^2} \\&= (r+2) \\&=(12)_r \end{align*}$$ Which is the RHS

So the equation is true for (almost) any $r$. But notice that in base $r$ we can express our number $(144)_r$ with a digit of $4$ (and indeed this is the highest digit we encounter in this problem in base $r$), so $r$ must be greater or equal to 5.

Answer 2

Simplify the expression first (remember that $r^0 =1$): $$\sqrt{r^2+4r+4}_r = (r+2)_r$$ $$\sqrt{(r+2)^2}_r = (r+2)_r$$ $$(r+2)_r = (r+2)_r$$

What's wrong with this statement: "This equality is true when $r ≥ -2$, because the square root of a real number is imaginary. Therefore, it is true for all (positive) bases."?

Hint: Look at the digits of the number in base $r$. For example, in base $2$, can we have the number $1323$?