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Divisibility (Posted on 2003-03-14) Difficulty: 4 of 5
In any set of 181 square integers, prove that one can always find a subset of 19 numbers, the sum of whose elements is divisible by 19.

  Submitted by Anoop    
Rating: 3.0000 (8 votes)
Solution: (Hide)
Consider the remainders obtained when the given 181 square integers are divided by 19. If n is any integer then n is congruent modulo 19 to one of the numbers 0, 1, 2, . . ., 18 and hence n^2 is congruent to 0, 1, 4, 9,16, 6, 17, 11, 7 or 5.

Hence there are only 10 possible values for the remainders. Since there are 181 remainders at least, one of them must repeat 19 times at least. Choose 19 numbers congruent to that remainder. This proves the result.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Solutionre: vague outlineMcWorter2005-04-02 19:54:29
Some Thoughtsvague outlineCharlie2003-03-18 03:48:29
re(2): conceptual solutionCharlie2003-03-18 03:27:52
Some Thoughtsre: conceptual solutionRavi Raja2003-03-17 04:24:32
re: conceptual solutionCharlie2003-03-17 03:58:40
conceptual solutionCory Taylor2003-03-16 07:38:49
re: Answer incompleteGamer2003-03-15 15:54:04
Answer incompleteTomM2003-03-15 06:18:45
SolutionI think I got itGamer2003-03-15 02:52:45
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