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? (For example, 256 and 530 have the common digit 5.)
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Anonymous
ANSWERS: 5
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There are a total of 10 digits. Each three digit integer uses three, so you need the next multiple of three greater than or equal to ten, which is 12 digits, or four three-digit numbers.
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Just two. I suggest you pick 256 and 530. Ok, ok, if you don't get to choose the numbers, then the worse thing to happen to you would be 999,888,777,666,555,444,333,222,111 none of which have any digits in common with each other. You can't be given any more numbers from 100-999 without using one of those digits again so ten numbers would be enough to guarantee a match.
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If you pick the right ones, then the minimum would be two. As long as you can pick which intergers, then just pick from among the ones with shared digits.
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10, you are unlucky to have nine picks 111, 222, 333 ... 999 with no common digits. The tenth has for sure a common digit.
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Since the sequence 111,222,333,444,555,666,777,888,999 satisfies the no digit in common constraint, the minimum number is at least 10.
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