Odd number of Birds
Can you put 46 birds into nine cages so that each cage contains an odd number of birds.
Labels: out-of-the-box, trickofmind
posted by Rajesh Lal at
7:59 AM
Tuesday, February 26, 2008
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20 Comments:
No.
Simple, no
if you take away one odd number, you are left with an odd number
if you take away two odd numbers, you are left with an even number
this shows that to leave 0 birds, you would need an even number of cages, which nine, obviously, is not
however this is only assuming that every bird must be caged, correct me if i'm wrong, but the question does not specifically say that all must be in a cage... though by reading it different ways it is possible to take this either way
agreed - no.
an odd number added an odd number of times equals and odd number, therefore 46 is not possible (or any other even number for that matter).
No one is thinking "inside the box... er cage" think of russian nesting dolls. You take a small cage and put one bird in it, you take a larger cage put 2 birds in it and the small cage in it. Now you have 43 uncaged birds, 7 empty cages, a cage with 3 birds in it and a cage with 1 bird... you can go from there, I say you put 7 bird in 4 of the cages, and 5 in the remaining 3... but really all this is up to you. All you really need is one cage within a cage.
L
That is a good idea. But I think we were thinking that we have nine similar cages.
That's the problem with sime puzzles, they aren't worded carefully enough. I am always very clear with my puzzle wording so there is only one possible answer. Except, of course, for funny answeres, which I love.
Cage inside cage i guess
as long as we're thinking outside the box, you can put two birds in, douse one with hot sauce, and only one remains. buwhahahaha.
as long as we're thinking outside the box, you can put two birds in, douse one with hot sauce, and only one remains. buwhahahaha.
no, because even/odd => even
example: 40/5=8
3516/3=1172
"Can you put 46 birds into 9 cages so that each cage contains an odd number of birds."
0 Is not odd, so every cage must have at least one bird, in theory.
If you set an odd number of birds in a single cage you then have eight cages left and X birds. However any two odd numbers make an even number, so you couldn't "nest" the cages; one cage would contain the next and so fourth end up with an even number from the cage within. Anytime you use more than one cage. So taking into account both math and physical birds you cannot put them in odd numbers throughout all cages.
However mathmatical -1 is still odd so you could have a lack of birds in one cage which would thearby balance out the '1 off' factor in the final equation.
In short, you cannot stack cages and reguardless of 'even seperation' or 'pyramid effects', you are still off by one, so you have to have an (or it was two if you are counting -1 as one) unused cage for others to be "proper".
Theoreticaly theory works.
Previous comments of nesting cages are correct. One cage with one bird is placed into another empty cage. This means two cages are used up and each contain one bird. Then a wide variety of options opens up with the remaining 45 birds and 7 cages. One option is 5 in one, 5 in one, 5 in one, 5 in one, 7 in one, 9 in one and 9 in one.
I wish I was this clever, but I heard a variation on this a long time ago:
Place one bird in each of eight cages. Then, in the last cage, you place the remaining thirty-eight.
Because when you think about it, 38is an odd number of birds to place in 1 cage.
The nested cage only counts if a cage inside a cage with one bird in each is considered odd (there are really 2 birds in the larger cage)
A much simpler answer is put 38 birds in one cage. Now one bird out in each of the other cages. Now let one bird go from the 38 (now 37).
Or put eight of the cages inside of each other with 45 birds in the smallest one. Now put one bird in the last cage.
chad your solution works..your math is off though... 1 bird in each of 8 cages then the remaining 37 birds in the last cage.(hope its a large cage)
His math is fine. 46 - 8 = 38.
if u dont have to have exactly the same amount of birds in each cage then u cloud have
46 birds 9 cages
1cage:7b=39b n 8 cages
2nd cage7b=32b 7c
3rd:7b=25birds 6c
4:7b=18b 5c
5:7b=11b 4c
6:3b=8b 3c
7:5b=3b2c
for the 8th cage but 1 bird cut the other in half but it in the cage, for the 9 cage but the 1 remaning bird then the other half of a bird in the cage lol
it not possibol
How many of the birds are carring fertile eggs?
Raj, plz could u post the answr thx!
Luke got it correct
i will post a number of asnwers by the weekend
-Raj
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