Adding 3, and still no change? Strange!
Number Theory
Level
5
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)
by
Aditya Raut