+0

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

0
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#1
+560
+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.

supermanaccz  Mar 21, 2018
#2
+65
0

thanks