# Adding 3, and still no change? Strange!

A number, when divided by 11, gives a remainder $$K$$. If the digit $$3$$ is placed at the end of the number, the remainder when divided by 11 still remains $$K$$.

Find the remainder when $$\displaystyle K^{(K-1)^{(K-2)^{\dots^{3^{2^1}}}}}$$ is divided by $$11$$.

Details and assumptions:-

The number is in base 10.

Placing the digit 5 at the end of $$123$$ makes the number into $$1235$$.

If $$K$$ was $$5$$, you had to find the remainder when $$5^{4^{3^{2^1}}}$$ is divided by $$11$$.

This is a part of the set 11≡ awesome (mod remainders)

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