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, 2015 at 20:53
  • $\begingroup$ Do you put the cards back? $\endgroup$ Apr 24, 2015 at 20:54
  • $\begingroup$ @VincentAdvocaat Riddles are mostly logic, logic is the base of mathematics and almost everything. $\endgroup$ Apr 24, 2015 at 21:11
  • $\begingroup$ logic puzzle, then $\endgroup$
    – smci
    Apr 25, 2015 at 0:07

1 Answer 1


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
    Apr 24, 2015 at 21:02
  • 2
    $\begingroup$ implicitly uses the Pigeonhole Principple $\endgroup$
    – smci
    Apr 25, 2015 at 0:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.