A box contains 900 cards enumerated from 100 to 999 (Each number appears once and just in one card). I took some random cards without looking at them and calculated the additions of the digits in each one.
How many cards should I take as minimum for making sure I have three cards with the same addition of digits?

  • $\begingroup$ Is this a riddle or a math problem? $\endgroup$ – Vincent Apr 24 '15 at 20:53
  • $\begingroup$ Do you put the cards back? $\endgroup$ – Ben Frankel Apr 24 '15 at 20:54
  • $\begingroup$ @VincentAdvocaat Riddles are mostly logic, logic is the base of mathematics and almost everything. $\endgroup$ – Juanvulcano Apr 24 '15 at 21:11
  • $\begingroup$ logic puzzle, then $\endgroup$ – smci Apr 25 '15 at 0:07

Assuming the cards aren't replaced, you would need to take

53 cards

In order to be sure that you have three cards where the digits add up to the same number.


There are 27 different possible sums, ranging from 1 (100) to 27 (999). Therefore, it would be possible to pull 27 cards without finding a single match. There are, however, only 25 repeatable sums, ranging from 2 (101/110) to 26 (998/899). There is only one combination of numbers that add up to 1, and only one combination that adds up to 27. Therefore, in order to get two of every repeatable sum and one of every non-repeatable sum, you would need 27+25 = 52 cards. There are no more possible duos, so pulling a 53rd card would guarantee that you had at least three cards with the same sum in their digits.

  • 1
    $\begingroup$ damn. just saw the question and thought immediately I knew how to do this. if I was a couple of minutes earlier I would have had the same answer $\endgroup$ – Ivo Beckers Apr 24 '15 at 21:02
  • 2
    $\begingroup$ implicitly uses the Pigeonhole Principple $\endgroup$ – smci Apr 25 '15 at 0:10

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