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What is the 5 numbers?

  1. Mar 17, 2006 #1
    1) imagine 5 numbers
    2) imagine the subsets of those 5 numbers.
    3) imagine the sum total of the elements in each subsets in 2).
    4) It is required that none of the sum of each subset are divisible by 5.
    5) What is the five numbers?
     
    Last edited: Mar 17, 2006
  2. jcsd
  3. Mar 18, 2006 #2
    What do you mean by number? Do you mean integers or any real number? I don't think this is possible with the integers if I am interpreting subset as I should be.
     
  4. Mar 18, 2006 #3
    Good question! It is integers
     
  5. Mar 18, 2006 #4
    But if the sum of the numbers in any subset of the 5 numbers can't be divisibe by 5, that means that no number in the original set can be divisible by 5, but then I'm pretty sure that no matter what you will end up with at least one subset that adds up to a number that is divisible by 5.
     
  6. Mar 18, 2006 #5
    I can see how it's possible with 4 numbers but not 5. I don't think it can be done, but hey, I also have to assume kant actually has a correct solution (OK, I hope he does!) The reason I don't think it can be done is because in Z5 (the field of integers modulo 5) there are only 4 equivalence classes that are non-zero.
     
  7. Mar 18, 2006 #6
    Yea that's waht I've been thinking, every integer has to be able to be expressed as one of

    5n
    5n + 1
    5n + 2
    5n + 3
    5n + 4

    for some integer n.

    5n is immediately out for any of the 5 numbers because the set of subsets includes just the single numbers so none of them can be divisible by 5. But then I can't think of any possibility using any of the others that will not give you some subset whose sum is divisible by 5.
     
  8. Mar 18, 2006 #7
    Clearly, as d_leet points out, none of the numbers can be divisible by 5. In fact, it is only necessary to consider the numbers modulo 5 since we only care about divisibility. Here are all the possible sets of 5 numbers modulo 5 and leaving out 0. The first 5 digits are the set (where the elements are arrayed in non-decreasing order), and the second set of digits is the subset whose sum is divisible by 5. It shows that the problem has no solution. x represents any digit.

    1. 11111 - 11111
    2. 11112 - 1112
    3. 11113 - 113
    4. 11114 - 14
    5. 1112x - 1112
    6. 1113x - 113
    7. 1114x - 14
    8. 1122x - 122
    9. 1123x - 23
    10. 1124x - 14
    11. 113xx - 113
    12. 114xx - 14
    13. 122xx - 122
    14. 123xx - 23
    15. 124xx - 14
    16. 1333x - 1333
    17. 1334x - 334
    18. 13444 - 3444
    19. 14xxx - 14
    20. 22222 - 22222
    21. 22223 - 23
    22. 22224 - 2224
    23. 2223x - 23
    24. 22244 - 244
    25. 223xx - 23
    26. 22444 - 244
    27. 23xxx - 23
    28. 244xx - 244
    29. 33333 - 33333
    30. 33334 - 334
    31. 1334x - 334
    32. 334xx - 334
    33. 3444x - 3444
    34. 44444 - 44444

    I wonder if kant means 'proper' subset in item 2. in that case, you can use {1, 1, 1, 1, 1}, {2, 2, 2, 2, 2}, {3, 3, 3, 3, 3} or {4, 4, 4, 4, 4}. If distinct integers are required, then use the fact that we only care about modulo 5. For example {3, 8, 13, 18, 23}.
     
    Last edited: Mar 18, 2006
  9. Mar 18, 2006 #8
    It would work if you could only use proper subsets of the 5 numbers. Then you could choose {n, n+5, n+10, n+15, n+20} as your set (assume n mod 5 not equal 0), and since 4n mod 5 is not zero, that works (which is how I got 4 numbers).
     
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