Find the least positive integer $n$ such that the two digits on the left of $n^{12}$ are equal Find the least positive integer $n$ such that the two digits on the left of $n^{12}$ are equal.
What I tried to find $n^{12}$ for $n=1,2,3,\dots,8$, but non of them was valid and it is tedious to raise integers to power $12$, the numbers are getting large rapidly.
@PredatorCorp used python, found that $n=18$.
Also I used MS-Excel: (Scientific notation does not matter here). So the solution is $n=18$.
But I am looking for a mathematical solution without any software.

Suggest me a hint to start solving this problem.
Your help would be appreciated. THANKS!
 A: The following function gives the first two digits of a positive integer written as $n^j$.
$$f(n,j)=\left\lfloor n^{j} 10^{2-\left\lceil \frac{j\log (n)}{\log (10)}\right\rceil }\right\rfloor$$
for $j=12$ and for $2 \le n\le 20$ gives
$$
\begin{array}{ccccccccccccccccccc}
 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\
 40 & 53 & 16 & 24 & 21 & 13 & 68 & 28 & 10 & 31 & 89 & 23 & 56 & 12 & 28 & 58 & 11 & 22 & 40 \\
\end{array}
$$
We can easily find that the lower integer such that $n^{17}$ has the first two digit equal is $n=8$
$$
\begin{array}{cc}
 2 & 13 \\
 3 & 12 \\
 4 & 17 \\
 5 & 76 \\
 6 & 16 \\
 7 & 23 \\
 8 & 22 \\
 9 & 16 \\
\end{array}
$$
A: Probably computing the powers directly is more efficient than using logarithms as in another answer.
If we compute $2^{12}=4096$ and $3^{12}=531441$, we can then prove that $18^{12}$ will have two matched digits on the left as follows:
$18^{12}=2^{12}×(3^{12})^2>(40×53^2)×10^6=(40×2809)×10^6=\color{blue}{11}2360×10^6$
$18^{12}=2^{12}×(3^{12})^2<(41×54^2)×10^6=(41×2916)×10^6=\color{blue}{11}9556×10^6$
This of course does not prove a minimal solution but does cap the number of trials we would need to establish minimality.
