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Krishnadas

  • 2 years ago

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?

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  1. ashwinjohn3
    • 2 years ago
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    If it were 10 rooms,the question would become easier.

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

  3. Krishnadas
    • 2 years ago
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    @ashwinjohn3 you are from mavelikkara?I am from Thrissur

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

  5. Krishnadas
    • 2 years ago
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    12th

  6. Krishnadas
    • 2 years ago
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    and you?

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

  8. ashwinjohn3
    • 2 years ago
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    11th

  9. Krishnadas
    • 2 years ago
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    @ashwinjohn3 no ...I am counting on self study

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

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

  12. wio
    • 2 years ago
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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.

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

  14. wio
    • 2 years ago
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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.

  15. Krishnadas
    • 2 years ago
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    @wio move??I didnt get that

  16. wio
    • 2 years ago
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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.

  17. Krishnadas
    • 2 years ago
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    its about accomodating...not moving them around

  18. wio
    • 2 years ago
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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.

  19. wio
    • 2 years ago
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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.

  20. Krishnadas
    • 2 years ago
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    but @wio each room should be accomodated

  21. wio
    • 2 years ago
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    Yeah, hold your horses and let me explain.

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

  23. Krishnadas
    • 2 years ago
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    ok

  24. wio
    • 2 years ago
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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.

  25. wio
    • 2 years ago
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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).

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

  27. wio
    • 2 years ago
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    Does that make sense so far?

  28. Krishnadas
    • 2 years ago
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    yes

  29. wio
    • 2 years ago
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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.

  30. wio
    • 2 years ago
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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

  31. wio
    • 2 years ago
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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.

  32. wio
    • 2 years ago
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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.

  33. Krishnadas
    • 2 years ago
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    @wio thanks

  34. wio
    • 2 years ago
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    Do you really understand it?

  35. wio
    • 2 years ago
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    Sometimes I explain things poorly.

  36. sauravshakya
    • 2 years ago
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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

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