2024/011) Find all values of n such that $6^n+1$ has all digits same.

We have $6^2=36$ that is it ends with 6.

So we have all the powers of 6 end with 6

So $6^n+1$ shall end with 7

So we need to find n such that all the digits of $6^n+1$ must have all digits 7 and let it be k 7's/

So $6^n+1 = \frac{7}{9} (10^k -1 )$

Or $9(6^n+1) = 7 (10^k -1 )$for

Or  $9 * 6^n + 16 = 7 * 10^k\cdots(1)$

We have $2 | 6$ so  $2^2 | 6^2 $ so if $n \ge  5$ then we have  

$9 * 6^n + 16 \equiv 16 \pmod {32} $ for $n \ge  5$

So   $9 * 6^n + 16 \equiv m  \pmod {32} $ where $m \le 16$ and not zero

For $k \ge 5$ $ 7 * 10^5 \equiv 0  \pmod {32} $

So $k \le 4$ 

So we need to check for n such that  $9 * 6^n + 16 \le = 70000$

Or $6^n \le  7776$

Or $6^n+1 \le 7777$

We calculate for n = 1 $6^1 + 1 = 7$ meets criteria

 n = 2 $6^2 + 1 = 37$ does not meet criteria

n = 3 $6^3 + 1 = 217$ does not meet criteria

n =4  $6^4 + 1 = 1297$ does not meet criteria

n = 5 $6^5 + 1 = 7777 $  meets criteria

Other value of n takes the value out of range

So $ n \in \{1,5\}$