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Krishnadas

I think this has no answer.. There are 50 horses and 9 rooms.Accomodate the horses in the rooms in such a way that no room contains even number of horses...Any idea?

  • one year ago
  • one year ago

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  1. ashwinjohn3
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    If it were 10 rooms,the question would become easier.

    • one year ago
  2. Krishnadas
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    @ashwinjohn3 if it was 10 i wouldnt have posted it here !lol

    • one year ago
  3. Krishnadas
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    @ashwinjohn3 you are from mavelikkara?I am from Thrissur

    • one year ago
  4. ashwinjohn3
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    @Krishnadas NICE!!!!!Finally a keralite in openstudy!!! in which class r u studying?

    • one year ago
  5. Krishnadas
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    12th

    • one year ago
  6. Krishnadas
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    and you?

    • one year ago
  7. ashwinjohn3
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    @Krishnadas going 2 pc thomas classes?

    • one year ago
  8. ashwinjohn3
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    11th

    • one year ago
  9. Krishnadas
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    @ashwinjohn3 no ...I am counting on self study

    • one year ago
  10. ashwinjohn3
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    @Krishnadas Well if u want IIT u must go to coaching classes

    • one year ago
  11. wio
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    I don't think it is solvable. I'm not sure how to prove it yet.

    • one year ago
  12. wio
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    Start by putting 1 horse horse in each room, and then the remaining 41 horses, in any room. You have 8 rooms with 1 horse and 1 room with 42 horses.

    • one year ago
  13. wio
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    Obviously in this setup, we fail because room 9 has an even number of horses.

    • one year ago
  14. wio
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    Now, from this set up, we could reach any other possible set up by moving a horse from any room to any other room.

    • one year ago
  15. Krishnadas
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    @wio move??I didnt get that

    • one year ago
  16. wio
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    I mean, we can reach any other possible set up by a series of 'steps' in which each step is moving a single horse from 1 room to another.

    • one year ago
  17. Krishnadas
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    its about accomodating...not moving them around

    • one year ago
  18. wio
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    I understand that. What I mean is this... suppose you start off by putting all of the horses in one room, got it? We can eventually get to any other accommodation by moving one horse at a time to some room.

    • one year ago
  19. wio
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    The motivation for this is to show there is an underlying property which CAN'T be violated regardless of your accommodation. Essentially it is proof by induction.

    • one year ago
  20. Krishnadas
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    but @wio each room should be accomodated

    • one year ago
  21. wio
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    Yeah, hold your horses and let me explain.

    • one year ago
  22. wio
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    Suppose you start out with 50 horses in one room and 0 in the rest. ok?

    • one year ago
  23. Krishnadas
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    ok

    • one year ago
  24. wio
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    You move a horse from one room to another. So you have a room with 49, a room with 1, and the rest have 0 horses.

    • one year ago
  25. wio
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    Whenever you move a horse between a room, both rooms change parity (change from even to odd or odd to even).

    • one year ago
  26. wio
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    This means that when you move a horse, you must change the parity of two rooms at a time.

    • one year ago
  27. wio
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    Does that make sense so far?

    • one year ago
  28. Krishnadas
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    yes

    • one year ago
  29. wio
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    There are four ways we can move a horse between rooms: 1) Move the horse from an odd room to an even one. In this case the number of even rooms stays the same, because the odd room becomes even and vise versa 2) Move the horse from an odd room to an odd one. In this case the number of even rooms increases by 2, because both rooms now have even number of horses. 3) Move the horse from an even room to an even one. In this case both rooms become odd, so the number of even rooms decreases by 2 4) Move the horse from an even room to an odd one. In this case, just like case 1, the number of even rooms stays the same.

    • one year ago
  30. wio
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    So, by moving a horse between rooms, the number of even rooms either: 1) doesn't change 2) increases by 2 3) decreases by 2

    • one year ago
  31. wio
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    So we start with 50 horses in one room. We have 9 even rooms. Your goal is to have 0 even rooms, but the number of even rooms must go down by 2, up by 2, or not change... so it's impossible to get 0 even rooms.

    • one year ago
  32. wio
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    Suppose someone came up with an accommodation that had 0 even rooms. We would be able to move every horse, one at a time, from the 8 other rooms into the 9th room. The number of even rooms could only change by 2 or 0 though, so it's impossible for us to end up with 9 even rooms... yet they would have to be able to do it... So basically, this is why it is impossible.

    • one year ago
  33. Krishnadas
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    @wio thanks

    • one year ago
  34. wio
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    Do you really understand it?

    • one year ago
  35. wio
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    Sometimes I explain things poorly.

    • one year ago
  36. sauravshakya
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    x1+x2+x3+x4+x5+x6+x7+x8+x9=50 Let x1,x2,x3,x4,x5,x6,x7,x8 be odd numbers then x1+x2 =even number x3+x4 =even number x5+x6 =even number x7+x8= even number And even number + even number = even number SO, x1+x2+x3+x4+x5+x6+x7+x8 = even number Now, Since 50 is even number x9 must also be even as x1+x2+x3+x4+x5+x6+x7+x8 is even. Thus, NO SOLUTION

    • one year ago
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