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26 Mar 2008 19:23:21 IST
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26 Mar 2008 21:22:03 IST
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oh sorry i didn't notice that u have to find within 1 day
actually my method was
to feed the pig the tonic bottle one by one for each hour
think we feed the first one at 1'o clock
second 2o clock.....
then calculate the time from 1'oclock to time of death
think its 34
34-24 = 10
its tenth bottle
actually my method was
to feed the pig the tonic bottle one by one for each hour
think we feed the first one at 1'o clock
second 2o clock.....
then calculate the time from 1'oclock to time of death
think its 34
34-24 = 10
its tenth bottle
26 Mar 2008 21:35:51 IST
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Wrong!!Answer is 7 pigs..
See,
how will you find one poisoned bottle out of seven bottles (numbered 1,2, 3,4,5,6,7).... take 3 pigs (unimaginatively named A, B and C) and feed them in such a way that there is a unique combination of deaths for each bottle that may be poisoned... eg. give 1,4,6 & 7 to pig A; 2,4,5 & 7 to pig B; 3,5,6 & 7 to pig C.... now, for every bottle, if it is poisoned, there is a unique combination of deaths:
if 1 is poisoned, only A dies
if 2 is poisoned, only B dies
if 3 is poisoned, only C dies
if 4 is poisoned, A & B die
if 5 is poisoned, B & C die
if 6 is poisoned, A & C die
if 7 is poisoned, A, B & C die
I give this example only to illustrate that the number of unique combinations possible between a given number of pigs is equal to the number of bottles that can be tested for poisoning at once.. in the above example, since 7 combinations were possible using A,B and C, 7 bottles could be tested using 3 pigs...
Please note that repetitions should not be considered..eg. AB is the same as BA here since both combinations point only to one bottle.. similarly, BC is same as CB... the point is although there are 6 two-lettered combinations possible between A,B and C (3*2*1 or 3! i.e 3 factorial), we can only consider 3 combinations (AB, BC, AC) as other 3 (BA, CB, CA)... this may seem trivial in case of three alphabets, but it will become increasingly important when the alphabets (or pigs) increase..
Now, what if there are 4 pigs A,B,C and D... how many combinations are possible.
4 single-lettered unique combinations i.e. A,B,C,D
double-lettered: totally, 12 combinations possible i.e. 4*3, but each combination is repeated twice like BA and AB.. so, only 6 useful unique combinations..
triple-lettered: totally, 24 combinations possible i.e. 4*3*2, but each combination is repeated 6 times like ABC can be written in 3! or 6 ways.. so, only 24/6= 4 unique combinations.
four-lettered: only one i.e. ABCD...
so, having 4 pigs will allow us to test 4+6+4+1= 15 bottles ... we try till this figure reaches 100 or more..
say, we have 5 pigs..
5 single-lettered combinations
20/2= 10 double-lettered combinations
60/6 = 10 triple-lettered combinations
120/24= 5 four-lettered combinations
1 five lettered combination.
so, 5 pigs let us test 5+10+10+5+1= 31 bottles
using similar calculations for 6 pigs, we get 63 combinations , which is still not enough....
get 7 pigs and we can test 7+21+35+35+21+7+1= 127 combinations ...
well, this is more than 100, but since 6 pigs will give us only 63 bottles, we need minimum 7 pigs to test for 100 bottles...
See,
how will you find one poisoned bottle out of seven bottles (numbered 1,2, 3,4,5,6,7).... take 3 pigs (unimaginatively named A, B and C) and feed them in such a way that there is a unique combination of deaths for each bottle that may be poisoned... eg. give 1,4,6 & 7 to pig A; 2,4,5 & 7 to pig B; 3,5,6 & 7 to pig C.... now, for every bottle, if it is poisoned, there is a unique combination of deaths:
if 1 is poisoned, only A dies
if 2 is poisoned, only B dies
if 3 is poisoned, only C dies
if 4 is poisoned, A & B die
if 5 is poisoned, B & C die
if 6 is poisoned, A & C die
if 7 is poisoned, A, B & C die
I give this example only to illustrate that the number of unique combinations possible between a given number of pigs is equal to the number of bottles that can be tested for poisoning at once.. in the above example, since 7 combinations were possible using A,B and C, 7 bottles could be tested using 3 pigs...
Please note that repetitions should not be considered..eg. AB is the same as BA here since both combinations point only to one bottle.. similarly, BC is same as CB... the point is although there are 6 two-lettered combinations possible between A,B and C (3*2*1 or 3! i.e 3 factorial), we can only consider 3 combinations (AB, BC, AC) as other 3 (BA, CB, CA)... this may seem trivial in case of three alphabets, but it will become increasingly important when the alphabets (or pigs) increase..
Now, what if there are 4 pigs A,B,C and D... how many combinations are possible.
4 single-lettered unique combinations i.e. A,B,C,D
double-lettered: totally, 12 combinations possible i.e. 4*3, but each combination is repeated twice like BA and AB.. so, only 6 useful unique combinations..
triple-lettered: totally, 24 combinations possible i.e. 4*3*2, but each combination is repeated 6 times like ABC can be written in 3! or 6 ways.. so, only 24/6= 4 unique combinations.
four-lettered: only one i.e. ABCD...
so, having 4 pigs will allow us to test 4+6+4+1= 15 bottles ... we try till this figure reaches 100 or more..
say, we have 5 pigs..
5 single-lettered combinations
20/2= 10 double-lettered combinations
60/6 = 10 triple-lettered combinations
120/24= 5 four-lettered combinations
1 five lettered combination.
so, 5 pigs let us test 5+10+10+5+1= 31 bottles
using similar calculations for 6 pigs, we get 63 combinations , which is still not enough....
get 7 pigs and we can test 7+21+35+35+21+7+1= 127 combinations ...
well, this is more than 100, but since 6 pigs will give us only 63 bottles, we need minimum 7 pigs to test for 100 bottles...
26 Mar 2008 21:41:56 IST
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There are two lengths of rope.
Each one can burn in exactly one hour.
They are not necessarily of the same length or width as each other.
They also are not of uniform width (may be wider in middle than on the end), thus burning half of the rope is not necessarily 1/2 hour.
By burning the ropes, how do you measure exactly 45 minutes worth of time?
Each one can burn in exactly one hour.
They are not necessarily of the same length or width as each other.
They also are not of uniform width (may be wider in middle than on the end), thus burning half of the rope is not necessarily 1/2 hour.
By burning the ropes, how do you measure exactly 45 minutes worth of time?
26 Mar 2008 21:49:04 IST
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Light both ends of rope 1 and one end of rope 2.
When the two burning portions of rope 1 meet, 30
minutes will have elapsed. Also, one-half of
rope 2 will have be burnt, meaning that there are
only thirty minutes of burning time remaining there.now light the other end of rope 2
when both the end of rope 2 meet now
15 min would elaps
hence
30 + 15 = 45
When the two burning portions of rope 1 meet, 30
minutes will have elapsed. Also, one-half of
rope 2 will have be burnt, meaning that there are
only thirty minutes of burning time remaining there.now light the other end of rope 2
when both the end of rope 2 meet now
15 min would elaps
hence
30 + 15 = 45
26 Mar 2008 21:57:16 IST
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Yes,u are right...another way is
burn one of dem from both the ends simultaneously... it'll burn fully in xactly 30 mins..
now fold d second one into half and again burn this from both d ends simultaneously...
it'll b completely burnt in xactly 15 mins..
so d total is xactly 45 mins..
burn one of dem from both the ends simultaneously... it'll burn fully in xactly 30 mins..
now fold d second one into half and again burn this from both d ends simultaneously...
it'll b completely burnt in xactly 15 mins..
so d total is xactly 45 mins..
29 Mar 2008 11:33:59 IST
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now my que is.....
There r 3 temples.and infront of the temples there r 3 wells...
so one guy brings some flowers wid him...the magic of the well is that what ever we put in it get doubles...and what we bring before entering into the temple we shud put them into the well.....when the boy enters into the temple...he should give some flowers to first temple..next again he should place the remained flowers into the next well and give same no of flowers to the 2nd temple as to the first...and procedure remains same to the third also....at the end of the temple there should not be any flowers with him...How much flowers did he bring and How much did he give to each of the temple???????
ex....
like if he bring 2 flowers wid him...it get doubles when he put in first well...so itz 4...so he gives 2 to first temple..now remained is 2 ..then he again put them into 2nd well again itz 4 and he gives 2 to temple..last case itz d same and 2 flowers remain...THIS SHOULD NOT HAPPEN...K...
There r 3 temples.and infront of the temples there r 3 wells...
so one guy brings some flowers wid him...the magic of the well is that what ever we put in it get doubles...and what we bring before entering into the temple we shud put them into the well.....when the boy enters into the temple...he should give some flowers to first temple..next again he should place the remained flowers into the next well and give same no of flowers to the 2nd temple as to the first...and procedure remains same to the third also....at the end of the temple there should not be any flowers with him...How much flowers did he bring and How much did he give to each of the temple???????
ex....
like if he bring 2 flowers wid him...it get doubles when he put in first well...so itz 4...so he gives 2 to first temple..now remained is 2 ..then he again put them into 2nd well again itz 4 and he gives 2 to temple..last case itz d same and 2 flowers remain...THIS SHOULD NOT HAPPEN...K...
29 Mar 2008 12:25:54 IST
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the boy takes 7 flowers with him
doubles them
7-----14
out of these 14 he offers 8
6 left
he again doubles them
6-----12
out of these 12 he offers 8
4 left
he again doubles them
4-----8
then he offers these 8
nothing left
so he should take 7 flowers with him
and in each turn offer 8
doubles them
7-----14
out of these 14 he offers 8
6 left
he again doubles them
6-----12
out of these 12 he offers 8
4 left
he again doubles them
4-----8
then he offers these 8
nothing left
so he should take 7 flowers with him
and in each turn offer 8
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I just have 1 day to find the poisoned bottle.