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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