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There is a number $x$ containing $1999$ digits and is divisible by $9$; if we add all digits of $x$ we get $y$ and as same adding digits of $y$ we get $z$. Find the probability that $z$ contains two digits.

My Attempt

let $y=9k \;; k\in \mathbb N$, $$y_{max}=9\times1999=17991 \qquad \qquad y_{min}=9 \qquad \qquad z_{max}=36(y=9999)$$ I have noticed one thing that for $k \in \{1,2,\dots,9 \}$ $z$ is not a double digit number.

For $k \in \{11,21,22,31,32,33,41,42,43,44,\dots,97,98,99\}$ ($45$ digits in total); $z$ is a two digit number. For rest all two digits values of $k$ number $z$ is a one digit number i.e. $9$.

For $k \in \{100, 120,130,140,150,160,170,190,200,230,240,250,260,\dots,800,890,900\}$ (i.e. Those 3 digit numbers which are $10's$ multiples of numbers not included in the privious set.) $z$ is not a two digit number. For rest all 3 digit vlues of $k$ $z$ is a two digit number.

For, $k\in \{1000,1200,1300,1400,1500,1600,1700,1800,1900\}$ ; $z$ is not two digit number.

So I get total $(9+45+45+9)=108$ values of $k$ for which $z$ is not a two digit number.

Probability=$\frac {1999-108}{1999}={1891\over1999} \approx 0.946$ but a very simple answer is given $2\over3$. Where am I wrong in approaching this question? Please help.

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    $\begingroup$ Your method is hard to follow. Knowing that $y\in \{9, \cdots, 17991\}$ there would appear to be $1999$ possible values of $y$, which of course need not occur with the same probability. You seem to only consider a handful of values for $y$ and you seem to assume they are all equally likely. $\endgroup$
    – lulu
    Commented Mar 17 at 15:24
  • $\begingroup$ Oh! Yes! I am getting your point but then how to handle this? My thought was simple. I did not consider that much. $\endgroup$
    – Skdmg
    Commented Mar 17 at 15:34
  • $\begingroup$ I would start by doing the same problem for a smaller number of digits, see if you can spot a pattern. Maybe it is useful to do $19$ and $199$, though it's not obvious to me that that's the best. I'd just try a bunch of cut offs. $\endgroup$
    – lulu
    Commented Mar 17 at 15:48
  • $\begingroup$ Main question I'd have is: how much does the probability vary? It's clear that $9\,|\,y$ but are there some values in the range that can't be hit? Are there others that get hit a lot? Experiment will tell you a lot. $\endgroup$
    – lulu
    Commented Mar 17 at 15:50

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