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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Can I choose 8 and 9? Then the sum is at most of two digit? – Jay Zha Mar 16 '17 at 03:46
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1I 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 $9$" – Stella Biderman Mar 16 '17 at 03:48
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Well, the sum can not exceed $9+9=18$. – Rayees Ahmad Mar 16 '17 at 03:48
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2I choose $e$ and $\pi$, with an infinite number of digits. – Joffan Mar 16 '17 at 03:48
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1@Joffan,sir, the question is," How namy digits does the interger have?" – Rayees Ahmad Mar 16 '17 at 03:55
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What a bizarre question. – Jonathan Hebert Mar 16 '17 at 04:29
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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 $2$.
Stella Biderman
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