Find the remainder of the following using binomial theorem?

Question

Find the remainder when $$ 5^{5^{5^{5...}}}$$ (24 times 5 ) is divided by 24 using binomial theorem ?

Answer to the question is 5 .

Answer 1

We have, for an integer $k\ge 1$, the following \begin{align*} 5^{4k+1}&=5\times(5^4)^{k}\\ &=5\times 625^{k}\\ &=5\left(24\times26+1\right)^k\\ &=5\left[(24\times26)^k+{k \choose 1}(24\times26)^{k-1}+\ldots+{k \choose k-1}(24\times26)+1\right]\\ &=5\times 24 \left[24^{k-1}\times26^k+{k \choose 1}\times 24^{k-2}\times 26^{k-1}+\ldots+{k \choose k-1}\times26\right]+5 \end{align*} Then, $5^5$ gives remainder $1$ under division by $4$, and so for $5^{5^5}$, $\quad5^{5^{5^5}},\quad\ldots$

Also, all these numbers give remainder $5$ under division by $24$.

Answer 2

$5^1=5, 5^2=25\equiv1\pmod{24}$.

So $5^{\rm odd}\equiv 5\pmod {24}$.

Now write your number as $5^x$, and check that $x$, being a power of $5$, is an odd number.

So the final answer is 5.