Consider: $$ y = \underbrace{19^{19^{\cdot^{\cdot^{\cdot^{19}}}}}}_{101 \text{ times}} $$ with the tower containing a hundred $ 19$s. Take the sum of the digits of the resulting number. Again, add the digits of this new number and get the sum. Keep doing this process till you reach a single-digit number. Find the number.

Here's what I tried so far:- Every number which is a power of $19 $ has to end in either $1 $ or $ 9$. Also, by playing around a bit, taking different powers of $19$, I found that regardless of the power of $19$, whether it is an odd or an even number, the single-digit number obtained at the end is always $ 1$. I've been trying to prove this, but I've no idea on how to do it. Can anyone help me out?

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    $\begingroup$ As $19\equiv1\pmod 9, 19^n\equiv1\pmod 9$ and among the single digits which is $\equiv1\pmod 9?$ $\endgroup$ – lab bhattacharjee Aug 5 '13 at 19:01
  • $\begingroup$ I'm sorry, but I do not understand. I'm new to modular arithmetic though I do understand what the notation means. How is the sum of the digits related to division by 9? $\endgroup$ – Train Heartnet Aug 5 '13 at 19:14
  • $\begingroup$ Repeatedly taking the digital sum (also known as a digital root) amounts to division by $9$. See this Wikpedia page: en.wikipedia.org/wiki/Digital_root $\endgroup$ – Adriano Aug 5 '13 at 19:19
  • $\begingroup$ @JobinIdiculla In the question you typed $100\text{ times}$, but then you say the tower contains one hundred $19s$. These options do not match. The former comprises of ninety nine $19s$. $\endgroup$ – Git Gud Aug 5 '13 at 19:25
  • $\begingroup$ @Git Gud: I meant a 100 towers of 19. Sorry, but my LaTeX knowledge is pretty limited and I'm not sure how to edit it. $\endgroup$ – Train Heartnet Aug 5 '13 at 19:49

$$10^0a_0+10^1a_1+10^2a_2+\ldots+10^na_n=(a_0+a_1+\ldots+a_n)+\text{a multiple of }9.$$

Therefore taking the sum of the digits of a number gives you a number that leaves the same remainder, in the division by $9$, as the one you had before (and it is also smaller as long as the original number is not $<10$). Therefore the process ends with a one-digit number that leaves the same remainder under division by $9$ are your number.


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