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The problem asks the following:

If $(\alpha2)_{r^3} = (r3\beta)_{r^2},$ find $\alpha, \beta,$ and r.

First off, I assume $r^2$ > 3 and $r^3 > 2$.

I know that $(r3\beta)_{r^2}$ = (10 _ _ _ _ )$_{r}$

I've attempted to convert to base 10 and set them equal to each other:

$(\alpha2)_{r^3} = \alpha ({r^3})^1 + 2({r^3})^0 $ = $ (\alpha {r^3} + 2)_{10}$

$(r3\beta)_{r^2} = r({r^2})^2 + 3({r^2})^1 + \beta ({r^2})^0 = (r({r^4}) + 3{r^2} + \beta)_{10}$

and converting to base r:

$\alpha = Mr^2 + Nr + O$

$ 2 = Xr^2 + Y$

$ r = 1r + 0$

$ 3 = Sr + T$

$ \beta = Vr + U$

But I'm pretty much stuck at this point. Don't see any logical progression from here (if I'm even headed in the right direction).

Any help is greatly appreciated.

Daniel Cazares
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1 Answers1

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Let: \begin{align*} \alpha &= Ar^2 + Br + C \\ 2 &= Dr^2 + Er + F \\ 3 &= Gr + H \\ \beta &= Ir + J \end{align*} where each uppercase coefficient is some integer in $\{0, 1, \ldots, r - 1\}$. Then by substituting into our expressions in base $10$, we have that: $$ (Ar^2 + Br + C)r^3 + (Dr^2 + Er + F) = r^5 + (Gr + H)r^2 + (Ir + J) $$ Comparing coefficients, we have: \begin{align*} \boxed{r^5}:\quad A &= 1 \\ \boxed{r^4}:\quad B &= 0 \\ \boxed{r^3}:\quad C &= G \\ \boxed{r^2}:\quad D &= H \\ \boxed{r^1}:\quad E &= I \\ \boxed{r^0}:\quad F &= J \\ \end{align*} Now let's assume that $r \geq 2$. Then $2 \geq Dr^2 \geq 4D$, which implies that $D = 0$. Hence, $H = 0$ so that $3 = Gr$. But then since $3$ is prime and $r \neq 1$, we have that $G = 1$ and $r = 3$.

Now since $G = 1$, we know that $C = 1$ so that $\alpha = 1(3)^2 + 0(3) + 1 = 10$.

Likewise, since $r = 3$, we know that $D = E = H = I = 0$ and $F = J = 2$ so that $\beta = 0(3) + 2 = 2$.

Adriano
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