+0

Suppose $(b_a)^2=71_a$, where $a$ and $b$ represent two distinct digits. If $b=a-1$, find $a$.

0
414
3
+211

Mar 21, 2018

#1
+638
+2

We can start by replacing b with (a-1).

\begin{align*} (a-1)^2\cdot a^0&=7\cdot a^1+1\cdot a^0
\end{align*}

So we can have 0 or 9.

But there is no base 0, so must be base 9.

Mar 21, 2018
#2
+211
0

thanks

Mar 22, 2018
#3
+638
0

Oh yay! My LaTeX finally worked for once.

Mar 30, 2018