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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?

Mathematics
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If it were 10 rooms,the question would become easier.
@ashwinjohn3 if it was 10 i wouldnt have posted it here !lol
@ashwinjohn3 you are from mavelikkara?I am from Thrissur

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Other answers:

@Krishnadas NICE!!!!!Finally a keralite in openstudy!!! in which class r u studying?
12th
and you?
@Krishnadas going 2 pc thomas classes?
11th
@ashwinjohn3 no ...I am counting on self study
@Krishnadas Well if u want IIT u must go to coaching classes
I don't think it is solvable. I'm not sure how to prove it yet.
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.
Obviously in this setup, we fail because room 9 has an even number of horses.
Now, from this set up, we could reach any other possible set up by moving a horse from any room to any other room.
@wio move??I didnt get that
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.
its about accomodating...not moving them around
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.
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.
but @wio each room should be accomodated
Yeah, hold your horses and let me explain.
Suppose you start out with 50 horses in one room and 0 in the rest. ok?
ok
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.
Whenever you move a horse between a room, both rooms change parity (change from even to odd or odd to even).
This means that when you move a horse, you must change the parity of two rooms at a time.
Does that make sense so far?
yes
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.
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
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.
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.
@wio thanks
Do you really understand it?
Sometimes I explain things poorly.
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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