2005 PMWC Problems/Problem I7
Problem
How many numbers are there in the list which contain exactly two consecutive 's such as and , but not or ?
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 $A \intersection B$ (Error compiling LaTeX. ) 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:
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 |