Adding 3, and still no change? Strange!

Let the number be $x$ . Thus we have $x \equiv K \pmod{11}$

Placing the digit $3$ at the end, makes the number $x$ into $10x+3$ .

Thus we have $10x+3 \equiv 10K+3 \equiv K \pmod{11}$ (The remainder hadn't changed, hence new remainder is same as $K$ )

This gives $10K+3 \equiv K \pmod{11} \\ \therefore 9K +3\equiv 0 \pmod{11} \\ \therefore 3K \equiv -1 \pmod{11} \\ \therefore 3K \times 7 \equiv -1\times 7 \pmod{11}\\ \therefore 21 K \equiv -K \equiv -7 \pmod{11} \\ \therefore K \equiv 7 \pmod{11}$

Hence $K=7$ .

Now we have to find the remainder when $7^{6^{...{2^1}}}$ is divided by $11$ .

By Fermat's little theorem,

$7^{10} \equiv 1 \pmod{11}$ And $6^{5^{...^{2^1}}} \equiv 6 \pmod{10}$ (The unit's digit of $6^k$ is always $6$ )

Hence $7^{6^{...{2^1}}} \equiv 7^6 \equiv 4\pmod{11}$

Hence the answer is $\boxed{4}$