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We choose two numbers between $0$ and $9$, and the sum of the numbers should not be $9$. In addition, we cannot choose a number which we have already taken. How many digits does the sum have at most?

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  • $\begingroup$ Can I choose 8 and 9? Then the sum is at most of two digit? $\endgroup$ – Yujie Zha Mar 16 '17 at 3:46
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    $\begingroup$ I have answered the asked question, but I wonder if it's what you intended. My answer has nothing to do with probability and doesn't use any of the assumptions of the problem other than "between $0$ and $9$" $\endgroup$ – Stella Biderman Mar 16 '17 at 3:48
  • $\begingroup$ Well, the sum can not exceed $9+9=18$. $\endgroup$ – Rayees Ahmad Mar 16 '17 at 3:48
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    $\begingroup$ I choose $e$ and $\pi$, with an infinite number of digits. $\endgroup$ – Joffan Mar 16 '17 at 3:48
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    $\begingroup$ @Joffan,sir, the question is," How namy digits does the interger have?" $\endgroup$ – Rayees Ahmad Mar 16 '17 at 3:55
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The sum of two one digit number is clearly less than $100$ and so cannot have $3$ digits. $6+5=11$ demonstrates that there can be $2$ digits, and so the answer is $2$.

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