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2005 PMWC Problems/Problem I7

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Problem

How many numbers are there in the list 1, 2, 3, 4, 5, \dots, 10000 which contain exactly two consecutive 9's such as 993, 1992 and 9929, but not 9295 or 1999?

Solution

We use casework:


Case 1: two digits


That's easy, one number.


Case 2: three digits

We can use PIE here:


Set A consists of all three digit numbers in the form 99a. 10 of those.

Set B consists of all three digit numbers in the form b99. 9 of those.

Set A \intersection B is the number 999.

There are 10+9-1=18 total three digit numbers. But we have a problem: we have counted 999, and that has 3 nines, not 2. So we subtract that to get

17 numbers here.


Case 3: 4 digits

So the numbers are either in the form ab99, a99b, and 99ab. We can just use direct counting here:

ab99: 9 choices for a (it can be 9, but no 0), and 9 choices for b (it can't be 9) =81. a99b: 8 choices (it can't be 0 or 9), and 9 choices for b (it can be 0 here) =72 99ab: 9 choices for a and 10 choices for b gives us 90.

72+81+90=243


we add all three cases:


1+17+243=\boxed{261}


See also

2005 PMWC (Problems)
Preceded by
Problem I6
Followed by
Problem I8
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10
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