- #1
jonroberts74
- 189
- 0
Homework Statement
How many integers from 100 through 999 must you pick in order to be sure that at least two of them have a digit in common? (they don't have to be in the same place value)
The Attempt at a Solution
worst case scenario involves picking integers such that none have had a digit in common
the other questions I've had like this have been all the possible combinations + 1 which is 9x10x10+1 = 901 but that doesn't seem correct.
seems like the only integers I could choose first that won't have a digit in common up to that point is 111,222,333,444,555,666,777,888,999 [assuming those are the first 9 integers I chose] then whatever I chose next will have a common digit with one of those.