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pythagoras123

Three different integers from 1 to 11 are selected. In how many of these combinations of 3 numbers are their sums a multiple of 3?

  • one year ago
  • one year ago

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  1. pythagoras123
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    how did you get 26? O_O

    • one year ago
  2. pythagoras123
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    erm... no.

    • one year ago
  3. inkyvoyd
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    Gah, FFM, you just gave me another strange problem to think about. The sum of the fib numbers?

    • one year ago
  4. dumbcow
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    the answer is 57 but i don't have a mathematical proof for you i used a computer to evaluate all possible cases

    • one year ago
  5. anonymoustwo44
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    @FoolForMath... Can I ask what combination of 3numbers is. does that mean order doesn't matter? or for instance, in this cause would 132 be same as 123?

    • one year ago
  6. FoolForMath
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    Btw what is the problem source?

    • one year ago
  7. pythagoras123
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    Homework

    • one year ago
  8. pythagoras123
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    That's why I need an explanation not just an answer

    • one year ago
  9. FoolForMath
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    Competitive math probably IMO?

    • one year ago
  10. pythagoras123
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    dunno, but looks like it

    • one year ago
  11. dumbcow
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    the distribution is as follows for combinations that sum up to 6,9,12 ... 1+3+7+11+13+11+7+3+1

    • one year ago
  12. FoolForMath
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    @anonymoustwo44: As this is combination 1+2+3 is same as 2+1+3

    • one year ago
  13. myko
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    you can try like that: the sum of numbers a,b,c is divisible by 3 if the sum of their digits is divisible by 3.

    • one year ago
  14. anonymoustwo44
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    @FoolForMath so the number 231 will be the same as the combination 321?

    • one year ago
  15. myko
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    they bouth divisible by 3

    • one year ago
  16. anonymoustwo44
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    so are they the same combination?

    • one year ago
  17. myko
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    what do you mean? the maximum sum to check in this case is: 11+10+9 = 30 (digits sum =3) you never even get to 123 or 321

    • one year ago
  18. anonymoustwo44
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    never mind me :))

    • one year ago
  19. Ishaan94
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    Every number can be expressed as 3n, 3n+1,3n+2 Case1. 3n,3n,3n Case2. 3n, 3n+1,3n+2 Case3. 3n+2,3n+2,3n+2 Case 4. 3n+1,3n+1,3n+1 In the set {1,2,..,11} 3 (3n forms), 4-4 (3n+1 and 3n+2) 3c3 + 3c1*4c1*4c1 + 4c3 + 4c1 How about this? :D

    • one year ago
  20. Ishaan94
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    3c3 + 3c1*4c1*4c1 + 4c3 + 4c3

    • one year ago
  21. Ishaan94
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    did i miss something? some case?maybe

    • one year ago
  22. Ishaan94
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    myko?

    • one year ago
  23. myko
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    just you have to be carefull with n=0.... wait checking...

    • one year ago
  24. Ishaan94
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    i only counted 3, for 3n forms {3,6,9}

    • one year ago
  25. Ishaan94
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    3n+1 3n+2 for n = 0 1 and 2

    • one year ago
  26. myko
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    never mind, you mean other thing, sry

    • one year ago
  27. myko
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    you right, good job

    • one year ago
  28. sheg
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    Ishaan is correct

    • one year ago
  29. sheg
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    Answer for this problem is 26

    • one year ago
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