+0

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

0
236
3
+195

Read the title

Rollingblade  Mar 21, 2018
#1
+627
+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
\\\Rightarrow\qquad a^2-2a+1&=7a+1
\\\Rightarrow\qquad a^2-9a&=0
\\\Rightarrow\qquad a(a-9)&=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
+195
0

thanks

Rollingblade  Mar 22, 2018
#3
+627
0

Oh yay! My LaTeX finally worked for once.

supermanaccz  Mar 30, 2018

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