100 Lives Are On The Line In This Week’s Puzzle. How Many Can You Save?

One - hundred prisoners all in a line . How many live and how many kick the bucket ? The answer depends wholly – well , almost entirely – on you .

This week ’s mystifier is one of my favorite twist on a popular panache of brain tease involve prisoners and hats . The destination of these brain-teaser is nigh always the same : A captive must get ahead her freedom – and , once in a while , the exemption of her fellow prisoner – by correctly identifying the color of her chapeau , which she can not see . How does a captive deduce the gloss of her own hat , base on the seeable hats and/or actions of her fellow prisoners ?

Two well - know magnetic variation on this genus of mystifier involvethree captives in a line , orfour prisoners arrange unmarried file , with the third and fourth prisoner separated by a paries ; many of you have suggested these brain teasers to me via e - post in recent week , but today ’s teaser involve significantly more prisoners – and a different approach – than either of these advantageously - eff versions . I ’ve borrow this interpretation fromtheTierney research lab postwhere I first encountered it . A few sentences have been modified for the sake of clarity .

Sunday Puzzle #9: The Puzzle Of 100 Hats

One hundred prisoners are lined up single file , face up in the same direction . Each prisoner will be randomly assigned either a flushed hat or a sorry lid . No one can see the color of his or her own hat . However , each person is capable to see the colouring material of the chapeau worn by every mortal in front of him or her . To wit , the person at the head of the line can not see the color of anyone ’s hat , the second prisoner can see only the first captive ’s hat , the third can see the first two prisoners ’ hat , and so on . The last person in strain – the 100th prisoner – can see the colors of the chapeau on all 99 citizenry in front of him or her .

Beginning with the last somebody in line , and then moving to the 99th person , the 98th , etc . , each will be asked to name the colour of his or her own hat . If the colour is correctly named , the individual exist ; if wrong named , the person is shot dead on the dapple . Everyone in ancestry is able to learn every answer as well as hear the gunshot ; also , everyone in line is able to remember all that needs to be remembered and is able to cypher all that need to be figure .

Before being lined up , the 100 prisoners are let to talk about strategy , with an eye toward developing a plan that will allow as many of them as possible to name the correct color of his or her own hat ( and thus survive ) . They know all of the preceding information in this problem . Once lined up , each person is leave only to say “ Red ” or “ Blue ” when his or her bend arrives , beginning with the last person in line .

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build up a plan that let as many multitude as possible to live . How many prisoner can you definitely save ?

Looking for a hint ? Here , in edict of plunderer - voltage , are some hints:1|2|3

We ’ll be back next week with the resolution – and a new puzzle ! Got a heavy brainteaser , original or otherwise , that you ’d wish to see featured?E - mail me with your recommendations.(Be sure to include “ Sunday Puzzle ” in the subject blood line . )

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Solution to Sunday Puzzle #8: A Monkey And His Uncle

Last week , I asked you all to take a crack ata devilishly phrase Son job . The mathematics , I tell you , was n’t terribly complex ( it requires basic algebra to solve ) , but I also monish that the puzzle would almost for sure involve an organized plan of attack to solve . After all , there ’s a stack of information to keep rails of in this puzzle – the weight of the scallywag , the weight of the uncle , the system of weights of the R-2 , and the rascal ’ historic period , for instance . And then , of course , there ’s that “ evil ” 5th sentence :

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The uncle is twice as old as the scamp was when the uncle was half as one-time as the monkey will be when the monkey is three times as sure-enough as the uncle was when the uncle was three times as old as the scalawag .

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Last week , I stated over and over again that a clear-cut plan of attack is the key fruit to clear many well logic puzzles . Something I did n’t mention , that was nonetheless masterfully demonstrated by several of you who relegate correct reply , is that a solution arrived at via a well - organise approaching is almost always easier to verify and explain than one bump upon by way of advert - hoc inquest . One of the most elegantly present solution to last week ’s puzzle come from Kevindrewster , whose response begins with a coherent explanation of how he undertake this time , which many of you visit as the crux of the puzzle :

I found it easiest to work this “ evil ” judgment of conviction in reverse . To commence , we roll in the hay that at some point in the past , the uncle was 3 time the age of the rascal . This does n’t give their ages , but it allows for a whole of measure to be show . Whatever age the scamp is at this time , let ’s call that 1 unit ( 1u ) . So , how old was the uncle when he was 3 times as old as the scalawag ? The uncle was 3u .

The old part of the sentence country that the monkey will someday be 3 meter this retiring age that we just established . So how old will the monkey be ? He will be 9u .

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Next , the uncle used to be half as old as this succeeding version of our scallywag . So , the uncle used to be 4.5u .

This first ( and sort - of final ) part of the sentence is the slippery . The uncle is currently twice as old as the MONKEY was when the UNCLE was 4.5u . Well , we know that they both senesce at the same pace , so we can extrapolate . If the scalawag was 1u when the uncle was 3u , then a twelvemonth and a one-half later when the uncle was 4.5u , the monkey must have been 2.5u . Thus , we know that the uncle is currently twice the age of 2.5u , making him 5u .

Now , we can figure out the monkey ’s current geezerhood using the same technique we just used . ( when the rapscallion was 1u , the uncle was 3u , so now that the uncle is 5u , the monkey must be 3u )

Ideapad3i

We also know that adding their ages will be 4 . So :

3u+5u=4

8u=4

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u=4/8

u=.5

This makes our monkey 1.5 years erstwhile , and his uncle 2.5

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Now , to cease off this mystifier ! The monkey ’s weight is the same as the uncle ’s age , so the monkey weighs 2.5lbs . And since they ’re suspend at equal distances , we jazz that they must weigh the same . This will help oneself us consider the circle

The final sentence of the puzzler is in reality pretty simple . The weight of the rope ( roentgen ) plus the weight of the uncle ( U ) is one - half again ( more simply put , 1 1/2 times ) the difference between the weightiness of the scallywag ( M ) and that of the uncle plus the monkey ( U+M ) .

R+U=1.5((U+M)-M )

Lenovo Ideapad 1

R+U=1.5U Then , take off atomic number 92 from both sides , and you get

R=.5U

So the rope is half the weight of the uncle , have it 1.25lbs , or 20 troy ounce . Knowing that each foot of rope is 4 oz. , we can found the distance of the rophy at 5 feet .

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I trust this was easy enough to translate . If nothing else , I get wind how hard it is to parse out complex algebraic word problem in a concise , leisurely to cover way .

Impressive chore ! Those looking for a purely mathematical answer to last workweek ’s riddle would do well to check up on outthis response from TristanW , which includes of a photo of some decent , unclouded , unionized equations .