# Homework Help: How many integers to pick such that 2 of them have digit in common

1. Jul 28, 2014

### jonroberts74

1. The problem statement, all variables and given/known data

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)

3. 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.

2. Jul 29, 2014

### Simon Bridge

Try finding 20 numbers from the range which do not have a digit in common.
Write them down.
That's right - so what is your conclusion?

3. Jul 29, 2014

### jonroberts74

In the worst case scenario, ten is the greatest number of integers I would have to choose in order to find another integer that had a digit in common with an already chosen integer because the first place value only has 9 choices to choose from and the tenth would cause a repeat of a digit. Assuming the other two place values have not had any shared digits--they would repeat on the 11 choice.

4. Jul 29, 2014

### haruspex

How can you find ten with no digits in common if the first position only has nine possibilities?

5. Jul 29, 2014

### jonroberts74

Not 10 without, the worst case senario would be the 10th choice would have a digit in common with the others because the first place value only has 9 choices.

Because I could somehow randomly choose

111
222
333
444
555
666
777
888
999

Then no matter what I choose next which is the tenth integer will have a common digit with one of the previous choices.

6. Jul 29, 2014

### haruspex

OK, I thought you were saying the answer was 11.

7. Jul 31, 2014

### Simon Bridge

... that does sort-of sound like you are saying that the answer is 11 doesn't it?

8. Jul 31, 2014

### Simon Bridge

... that does sort-of sound like you are saying that the answer is 11 doesn't it?

9. Jul 31, 2014

### vela

Staff Emeritus
I was initially confused by that bit at the end too, but all the OP meant was that the second and third digits would definitely be a repeat on the 11th number because there are 10 choices for those positions.

10. Jun 17, 2015

### Mitchel Haas

There are 9*9*8 = 648 integers which do not have a repeating digit.
(1 - 9) * (0 - 9 -1) * (0 - 9 - 2)
So, you must pick 649 to insure you have an integer with a repeating digit.

11. Jun 17, 2015

### haruspex

Indeed, but that is not the question in this thread.