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How many integers to pick such that 2 of them have digit in common

  1. Jul 28, 2014 #1
    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. jcsd
  3. Jul 29, 2014 #2

    Simon Bridge

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    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?
     
  4. Jul 29, 2014 #3
    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.
     
  5. Jul 29, 2014 #4

    haruspex

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    How can you find ten with no digits in common if the first position only has nine possibilities?
     
  6. Jul 29, 2014 #5

    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.
     
  7. Jul 29, 2014 #6

    haruspex

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    OK, I thought you were saying the answer was 11.
     
  8. Jul 31, 2014 #7

    Simon Bridge

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    ... that does sort-of sound like you are saying that the answer is 11 doesn't it?
    So you have your answer?
     
  9. Jul 31, 2014 #8

    Simon Bridge

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    ... that does sort-of sound like you are saying that the answer is 11 doesn't it?
    So you have your answer?
     
  10. Jul 31, 2014 #9

    vela

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    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.
     
  11. Jun 17, 2015 #10

    Mitchel Haas

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    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.
     
  12. Jun 17, 2015 #11

    haruspex

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