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I’m doing a review exercise that gives me the list of numbers from 100 to 1000.

I need to find the number of different numbers that have a 0.

I suppose I could do this with the Pigeonhole principle, but I’m not sure how to implement it.


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Wouldn't it be more natural just to count them? $100,101,\dots,109$ all have a $0$, as do $110,120,\dots 190$. That's $19$ numbers with a $0$ in $100-199$, and so on. – yunone Jun 17 '11 at 22:24
Hmm, then maybe I can approach this using Induction. – Matt Jun 17 '11 at 22:33
I don't think induction is really necessary here. That's more useful for when you want to prove some property for all natural numbers. Here the case is simple. There are $19$ numbers with a $0$ between $100-199$. The same holds for $200-299$, $300-399$, etc. up to $900-999$, as it doesn't matter what the hundreds digit is, as long as it's not $0$. You've exhausted your whole list then, and thus counted all the numbers with a $0$. – yunone Jun 17 '11 at 22:37
At least one zero or exactly one zero? For "at least" I would find the number $n$ of numbers $100$ to $999$ that have no zero. This is easy, turns out that $n=729$. There are $900$ numbers $100$ to $999$, so $171$ with at least one zero. Finally, let's remember about $1000$, which has been sadly neglected. So we get an answer of $172$. If you want instead exactly one zero, don't need to bother about $1000$, adjust the $171$ at the end to remove the two zero people. – André Nicolas Jun 17 '11 at 23:11

I don't really know how you would do this with the pidgeonhole principle. I would count how many numbers have no zeros, then how many numbers have 2 zeros (hint, it's 9), then use that to find how many numbers have 1 zero. Then you can use all of that information to get the total number of zeros.

There's a common theme when counting that it's sometimes easier to count everything except what you're asked to count and then subtract.

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