my favorite logic puzzles

8/19/2019 My Favorite Logic Puzzles

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My Favorite Logic Puzzles

by John P. Prattlast updated 17 Feb 2007

None of the following puzzles have tric answers! or unwarranted assu"ptions. #heyare arranged appro$i"ately in order of difficulty% the first few can be done byele"entary school children! but they "ay need to be taught how to thin. & tried toinclude a wide variety of types of puzzles! so that each would teach a new lesson ineither logical thining or finding creative solutions by breaing a variety of inds of"ental blocs. 'oreover! "ost of the" can be solved "entally by using a (udicious

point of view. )in to answers is found at the end! but don*t go there until you*vesweat plenty+

1. ,rown! Jones and -"ith are a doctor! a lawyer! and a teacher. #he teacher! whois an only child! earns the least "oney. -"ith! who "arried ,rown*s sister!earns "ore than the lawyer. hat is each "an*s (ob/

2. far"er has a fo$! goose and a bag of grain! and one boat to cross a strea"!which is only big enough to tae one of the three across with hi" at a ti"e. &fleft alone together! the fo$ would eat the goose and the goose would eat thegrain. ow can the far"er get all three across the strea"/

. drawer contains 10 blac and 10 brown socs that are all "i$ed up. hat isthe fewest nu"ber of socs you can tae fro" the drawer without looing and

be sure to get a pair of the sa"e color/

3. #here are three bo$es which each contains two "arbles4 one has two white! onehas two blac and one has one white and one blac "arble. 5ach of the bo$esalso is labeled as to its contents! but each label is incorrect. hat is the fewestnu"ber of "arbles you could re"ove fro" the bo$es and loo at in order todefinitely deter"ine the contents of all three bo$es/

6. hen ased her children*s ages! 'rs. 'uddled said that lice is the youngestunless ,ill is! and that if arl isn*t the youngest then lice is the oldest. ho isthe oldest and who is the youngest/

8. #wo fathers and two sons went fishing. 5ach caught e$actly one fish and yetthere were only three fish caught. hy/

7. 'r. and 'rs. -"ith and their two children for" a typical "erican fa"ily. &fe$actly two of the following state"ents are true! what is the first na"e of thefather! "other! son and daughter4


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o 9eorge and :orothy are blood relatives.

o oward is older than 9eorge.

o ;irginia is younger than oward.

o ;irginia is older than :orothy.


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nows that the native could be either fro" the tribe of #ruthBtellers! who alwaystell the truth! or fro" the )iars! who always lie. hat is one @uestion he couldas the native to discover the path to the village/

16.#welve billiard balls all weigh the sa"e! e$cept for one that is either light or

heavy. :eter"ine in three weighings on a balance scale which is the Codd ballC!and whether it is light or heavy.

18. prisoner was given a chance to be blindfolded and pic one ball fro" two bowls that would contain a total of 60 white and 60 blac balls. hoosing white"eant freedo"! blac "eant death. e ased if he could divide the balls

between the bowls before he was blindfolded and his re@uest was granted.hat is the best way to divide the balls between the bowls/

17.ow can three "issionaries and three cannibals cross a river in a canoe thatholds at "ost two people if the cannibals "ust never outnu"ber the"issionaries on either side/

1.#he following puzzle is about two actual historial people.

9eorge and 5velyn never "et but they carried on writing until late in life. &t has been said that 5velyn loved 9eorge! but she was too old for hi". 9eorge"arried in 1


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ANSWERS

1. ,rown! Jones and -"ith are a doctor! a lawyer! and a teacher. #he teacher! who

is an only child! earns the least "oney. -"ith! who "arried ,rown*s sister!earns "ore than the lawyer. hat is each "an*s (ob/

-"ith earns "ore than the lawyer and also "ore than the teacher Dwho earnsthe leastE so -"ith "ust be the doctor. ,rown has a sister but the teacher hasnone! so ,rown "ust be the lawyer. #hat "eans Jones "ust be the teacher.

=ne way to solve al"ost any of this whole class of proble"s Dwhich can gete$tre"ely co"plicatedE is to "ae a grid with the na"es in one direction andthe occupations in the other. #hen one by one eli"inate cells in each row orcolu"n fro" the facts given. hen all but one of a row or colu"n is filled in!then fill in the rest of the corresponding row and then continue. & have seenwhole boos of (ust these inds of proble"s! which see"s boring to "e

because once one learns this techni@ue! they can all be solved the sa"e way. &only include one of these proble"s and this is it. )et*s do this proble" to seehow it wors.

For e$a"ple! start with the following table4

:octor )awyer #eacher

,rownJones

-"ith

#hen "ar an in i"possible cells. -"ith cannot be the lawyer! nor theteacher.


,rown

Jones-"ith

#hen! when only one possible cell re"ains in a row or colu"n! "ar an = thereand then fill in the rest of the re"aining colu"n or row. #hat is! once we nowthat -"ith is the doctor! we now that ,rown and Jones are not the doctor.


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,rown

Jones

-"ith =

Now continue crossing off i"possible cells. ,rown cannot be the teacher4


,rown

Jones

-"ith =

Now that only one cell re"ains possible in both row one and colu"n three!

each "ust be an =.


,rown =

Jones =

-"ith =

#he puzzle is now solved because there is an = in each row and colu"n. &f youreally feel the need to! you "ay place an in the "iddle cell! but the puzzle is

already solved at this point. #hat is! fro" the table you can see that ,rown isthe lawyer! Jones is the teacher! and -"ith the doctor. #his "ethod "ight see"lie overill for this si"ple proble"! but once you learn how to draw thesegrids! you can solve the hard puzzles! lie identifying the positions of ninena"ed players on a baseball tea". )et*s "ove on to another whole class of

proble".

2. far"er has a fo$! goose and a bag of grain! and one boat to cross a strea"!which is only big enough to tae one of the three across with hi" at a ti"e. &fleft alone together! the fo$ would eat the goose and the goose would eat the

grain. ow can the far"er get all three across the strea"/

,efore & try to solve any proble" & usually draw a picture and even though &don*t here! & did when & solved it Dhighly reco""endedE. #he first trip re@uiresthe far"er to tae the goose across! because it cannot be left with either grain or the fo$. =n the ne$t trip! the far"er "ay tae either fo$ or grain! but then he"ust bring bac the goose. =n the third trip! the far"er taes the re"aining


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ite"! but leaves the goose on the first shore. e then "aes a fourth trip to bring the goose. #hus! four trips are re@uired! and the goose ends up crossingthree ti"es.

. drawer contains 10 blac and 10 brown socs that are all "i$ed up. hat is

the fewest nu"ber of socs you can tae fro" the drawer without looing and be sure to get a pair of the sa"e color/

&n this proble"! it is very i"portant to notice that the color re@uired is notspecified. &t is only necessary to pic three socs to be guaranteed of getting atleast two blac or two brown socs.

3. #here are three bo$es which each contains two "arbles4 one has two white! onehas two blac and one has one white and one blac "arble. 5ach of the bo$esalso is labeled as to its contents! but each label is incorrect. hat is the fewest

nu"ber of "arbles you could re"ove fro" the bo$es and loo at in order todefinitely deter"ine the contents of all three bo$es/

-olving this proble" again involves reading it very carefully and nowing thatit "eans (ust what it says. e are told e$actly what is in each bo$ and also thatall three labels are wrong Dnot (ust possibly wrongE. #hat is so a lot ofinfor"ation! so let*s draw a grid as in proble" 1! crossing off the possibilitiesthat any bo$ is label correctly4

)abeled ,B, )abeled ,B )abeled B

ontains ,B,

ontains ,B

ontains B

#his proble" now differs fro" the first in that we "ust choose (udiciously tofill in the rest of the table in the fewest nu"ber of loos inside the bo$es. #he

best way & now to discover the answer is to loo for a choice where you areguaranteed to learn so"ething new. &f you loo first in the bo$ labeled ,B, andyou pull out a white "arble! you still haven*t learned what that bo$ contains

because it "ight be ,B or B. -i"ilarly if you loo in the B bo$! achoice of , does not allow you even to fill in one "ore cell of the table. ,ut ifyou choose the ,B bo$! then you are guaranteed to be able to fill in one "orecell! because if you draw out a white "arble! then that bo$ "ust be the B. &fyou draw out a blac! then it "ust be the ,B,. )et*s suppose it is white! and fillin the table4


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ontains ,B,

ontains ,B

ontains B =

Now we can fill in the rest of colu"n two and row three because only one cellis left4


ontains ,B,

ontains ,B

ontains B =

nd now! wonder of wonders! the first colu"n only has one e"pty cell! as doesthe first row! so we can fill the" in too4


ontains ,B, =

ontains ,B =

ontains B =

-o the a"azing thing is that by drawing only one "arble out of the bo$ labeled

,B! we can deduce what is in all three bo$es+

&f we had been bright enough to solve it without the table we could have saidthat the ,B bo$ "ust be the B because we drew a white "arble out of it.#hat would "ean the bo$ labeled ,B, "ust contain the ,B because it is"islabeled and cannot contain two blac "arbles. #hat leaves the bo$ labeledB to contain two blac "arbles. #hus! the puzzle can be solved withoutusing a grid! but to "e the grid "ethod is "ore convincing because the logic isso clear at each step.

6. hen ased her children*s ages! 'rs. 'uddled said that lice is the youngestunless ,ill is! and that if arl isn*t the youngest then lice is the oldest. ho isthe oldest and who is the youngest/

gain! let*s resort to the grid "ethod! because wors so well. &f you are alreadygetting bored with it! note that every one of these proble"s introduces a new


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twist. #his one is that 'rs. 'uddled is confusing us with state"ents with theword CunlessC in it. ow do we "ar that in a grid/ First! let*s set up the grid.

lice ,ill arl

=ldest'iddle

Goungest

=ay to start! we are told that the youngest "ay be either lice or ,ill. owcan we use that infor"ation/ ell! we can deduce that arl cannot theyoungest! so we can cross off that cell4

lice ,ill arl

=ldest

'iddle

Goungest

Ne$t we are told that if arl isn*t the youngest Dwhich he*s notE the lice is theoldest. -o lice "ust be the oldest and we can fill in that bo$4

lice ,ill arl

=ldest =

'iddle

Goungest

Now we can fill in the rest of the first colu"n and first row because no one elsecan be oldest and lice cannot be anything but oldest4

lice ,ill arl

=ldest =

'iddle

Goungest

Finally we can fill row three and colu"n three which both have only onere"aining cell available4

lice ,ill arl

=ldest =


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'iddle =

Goungest =

#hus! lice is oldest! arl is 'iddle and ,ill is the youngest.

8. #wo fathers and two sons went fishing. 5ach caught e$actly one fish and yetthere were only three fish caught. hy/

#his is a classic in creative thining. 'any proble"s are set up to get you toe$pect one thing and then shoc you be having the obvious be false. t first wethin that two fathers and two sons "eans four people. ,ut each caught a fishand yet only three fish were caught. ence! there were only three people. owcould that be/ &t "ust be that one of the three people fulfills a dual role of being

both a father and a son. &s that possible/ =f course. -o it "ust have been a

grandfather! a father and his son who went fishing.

7. 'r. and 'rs. -"ith and their two children for" a typical "erican fa"ily. &fe$actly two of the following state"ents are true! what is the first na"e of thefather! "other! son and daughter4

o 9eorge and :orothy are blood relatives.

o oward is older than 9eorge.

o ;irginia is younger than oward.

o ;irginia is older than :orothy.

e could draw a 3$3 grid to solve this but it "ay not be a great idea becausewe don*t even now which state"ents are true. ,esides! all we need to find outis whether the father is 9eorge or oward! and whether the "other is :orothyor ;irginia. e "ight be as well off with two 2$2 grids! one for the "ales andone for the fe"ales. ,ut the spirit of this proble" is "ore to deter"ine whichstate"ents are true and which false! so let*s try doing this one without grids.

e could try starting with each pair of state"ents and testing to see whether if

two are true! are the others necessarily false. =r we could (ust guess each of thefour answers in turn to see which one wors. ,ecause other si"ilar proble"s"ight have too "any solutions to try! let*s do it the first way. #he secret todoing the proble"s with so"e true and so"e false state"ents is to try totally

believing a state"ent and deter"ining whether it leads to a logicalcontradiction.


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First let*s try believing the first two state"ents to be true and see if that wors.-tate"ent H1 "eans that 9eorge and :orothy are not both parents! H2 "eansthat oward is that father. -o far so good% now we need the last two state"entsto be false. H could be false only if ;irginia is the "other! so that would "eanoward and ;irginia are the parents. ,ut then H3 would also be true! so this is

not the solution because that would be true state"ents.

-o now let*s suppose H1 is false. #hen 9eorge and :orothy are the parents. #hiswould "ean that H2 is also false because 9eorge would be older than his sonoward. an the last two state"ents be true/ -tate"ent H could be true

because ;irginia could be oward*s younger sister. ,ut! alas! H3 would befalse! which is too "any false state"ents.

#hus we have deduced that H1 "ust be true and H2 "ust be false if this puzzleis to have any solution at all. -tate"ent H2 false "eans that 9eorge is that

father! and H1 true then "eans that ;irginia is the "other. #hat is the only possible solution left! so it had better wor. -tate"ent H "ust then be false because ;irginia "ust be older than her son oward. #hat "eans H3 needs to be true! and G5-! it is4 ;irginia would be older than her daughter :orothy. -othere is e$actly one solution4 9eorge and ;irginia are the parents of owardand :orothy.


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front were not both red! that is! at least one of the" was blue. 'ost people getthis far in the puzzle and are then stu"ped because they thin there is no way totell which one is blue! or perhaps both. #he tric to solving the puzzle is to now

put yourself in the place of the second "an. &f you were in his shoes and heardthe first answer! you would also now that at least one of that front two hat was

blue. &f you saw a red hat in front of you! you would now you had a blue hat.,ut he did not now. #he only thing that could have prevented hi" fro"nowing is that the first hat "ust have been blue.

>. hile a red "ar was placed on the forehead of each of three blindfoldedwo"en seated facing each other in a circle! they were told that the the "ar"ight be either red or white. ?pon re"oval of the blindfolds! each was to raiseher hand if she saw at least one red "ar! and then to tae it down if she couldlogically deduce the color of her own "ar. ll three hands were @uiclyraised! but then one of the" lowered her hand. ow did she now/

#his is a variation of the above proble"! but very hard for so"e people tosolve. 5veryone can see that each wo"an could be thining that her ownforehead "ight be red or white because the other two wo"en could be raisingtheir hands because of each other. gain! this is the point where "ost peoplestop! and indeed! where the wo"en stopped while all three hands were up. #hetric to solving this proble" is to A5))G put yourself into the s"art wo"an*sshoes. &f you A5))G were she! you*d say! either & have a white spot or a red.-uppose you had a white spot. #hen the other two wo"en would be looing atone white spot and one red. #hey would each @uicly figure out that that onlyreason the others hand was up was because of their own red spot. #he fact thatneither of the" figured it out was the tip off to the first that she "ust also havea red spot. 'ost people put the"selves in the place of the first person! but tosolve this one! you "ust then also put yourself in the place of a second wo"an.

10.'r. Aeader*s five daughters each gave boos for hrist"as to one or "ore ofher sisters. 5ach presented four boos and each received four boos! but no twogirls allocated her boos in the sa"e way. #hat is! only one gave two boos toone sister and two to another. ,eth gave all her boos to lice% hristy gavethree to 5dith. hich sisters gave the four boos to :eborah/

#his sounds lie a perfect one for the grid "ethod! but we*re going to fill thegrid with the nu"ber of boos given to each sister. Note that it is not re@uiredto tell who gave how "any boos to each sister! but only to :eborah. &t "aynot be solvable all the way! but let*s try. First let*s "ae a grid with what isgiven! with na"es at left "eaning C9ivesC and na"es at top "eaningCAeceivesC.


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lice ,eth hristy :eborah 5dith

lice 0

,eth 3 0

hristy 0

:eborah 0

5dith 0

=ay! now what/ #here are a lot of cells to fill in. ow do we start/ First! wecan see that lice already received her four! so no other sister gave her any! sowe can put zeros in the rest of the first colu"n! and also in the rest of her row.'oreover! we can notice that one sister gave one boo to each of her sisters.#hat sister had to be lice! because none of the other other sisters could givelice a boo because she had received her four already. #hus we can fill in the

first row too4


lice 0 1 1 1 1

,eth 3 0 0 0 0

hristy 0 0

:eborah 0 0

5dith 0 0

Now we loo at the total in the last row and see that it is already 3. -o the lastcell in the last row "ust be zero4


lice 0 1 1 1 1

,eth 3 0 0 0 0

hristy 0 0

:eborah 0 0 0

5dith 0 0

Now we co"e to the proble" that we don*t now if hristy gave her fourth boo to ,eth or :eborah! and we need to now that to solve the proble". ,utthere is a way to eep filling in the table other ways. )oo at :eborah*s boosreceived. -he only has one so far and needs to get four and has only two sistersleft to give the" to her. #hus! she "ust get one boo fro" one sister and two


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fro" the other. ,ut :eborah cannot receive two fro" hristy! who only has oneleft to give. -o :eborah "ust receive one fro" hristy and two fro" 5dith4


lice 0 1 1 1 1,eth 3 0 0 0 0

hristy 0 0 1

:eborah 0 0 0

5dith 0 2 0

#echnically we are done at this point because we can say that :eborah receivedone boo fro" lice and hristy! and two fro" 5dith. ,ut how can we resistfilling in the rest of the table/ learly :eborah "ust be the one giving away

two boos to each of two sisters because there are only two left to give the" to.nd so 5dith "ust the be sister who split up the boos 2B1B1. #hus we can fillin the entire table. Note that each gives four boos and each receives four4


lice 0 1 1 1 1

,eth 3 0 0 0 0

hristy 0 0 0 1

:eborah 0 2 2 0 0

5dith 0 1 1 2 0

&sn*t that an astounding a"ount of infor"ation to fill in fro" so little given/#hat is why this proble" is one of "y favorites.

11.ow can you connect nine dots in three straight rows of three with four straightlines without raising the pencil fro" the paper/

#his calls for a creative answer. &t re@uires going beyond the end of the s@uareof nine dots4

O------O-------O

|\ /

| \ /

| \ /

O O O

| \ /

| \ /

| /

| / \


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O O O

| /

| /

| /

| /

|/

12. "an needed to pay his rent and was out of "oney! but found that his rent wasworth about one gold lin on his chain per day. hat is the fewest nu"ber ofcuts he can "ae in his 2Blin chain to pay the rent for up to 2 days/

#his proble" is great because it re@uires only two lins to be cut. ut linnu"ber 3 and lin nu"ber 11 counting fro" the sa"e beginning lin. e thenhas 2 pieces of length 1 Dthe cut linsE! and one of ! 8! and 12. e can then paythe rent as follows. =ne each of the first two days he can give a cut lin. =n thethird day he gives the chain of and gets his two cut lins bac. e uses the"on days 3 and 6! and then trades all given so far and gives the 8Blin chain on

day 8. e then again repeats the first steps for days 7B11. =n day 12 he gets allthose lins bac and gives the 12Blin chain. #he then repeats the actions of thefirst 11 days to go all the way though day 2. For those nowing nu"beringsyste"s! it will be noticed that this is basically a trinary nu"bering sche"e.

1.n e$plorer wishes to cross a barren desert that re@uires 8 days to cross! butone "an can only carry enough food for 3 days. hat is the fewest nu"ber ofother "en re@uired to help carry enough food for hi" to cross/

#here are probably a lot of ways to solve this but "y way was first to realize

that if each of the "en ate his own food then even if we begin with a hundred"en! each can only get four days into the desert. learly the idea is to get onlythe one e$plorer across and have the helpers return bac. ,y si"ply trying afew ideas the answer is clearly that two other "en are re@uired.

#he first helper only goes on day into the desert. e feeds the other two "enduring the first day! so that at the beginning of the second day! he only has oneday rations left. -o he goes bac to ca"p. =n the second day! the second helper feeds hi"self and the e$plorer. =n the beginning of the third day the helpernow has two days rations left so he heads bac. #he e$plorer is two days into

the (ourney and still has all four days of his food left! so he continues on alone.

13. traveler "eets a native in the (ungle at a the for in the trail! where only one path goes to the village. #he traveler does not now which path to tae andnows that the native could be either fro" the tribe of #ruthBtellers! who alwaystell the truth! or fro" the )iars! who always lie. hat is one @uestion he couldas the native to discover the path to the village/


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#here are at least a hundred variations on this proble" and they are not all based on the sa"e idea. #his is one of the si"plest but it is still tricy. #he ideais to find a @uestion such that either native will answer the sa"e. #hat re@uiresthat the @uestion include so"e sort of double negative! which will cancel outthe fact that he "ay be lying. =ne e$a"ple is C&f you were fro" the other tribe!

which path would you say goes to the village/C #he truth teller will point to thewrong path! and so will the liar Dthin about itE. -o whatever path is pointed to!tae the other one.

16.#welve billiard balls all weigh the sa"e! e$cept for one that is either light orheavy. :eter"ine in three weighings on a balance scale which is the Codd ballC!and whether it is light or heavy.

hat "aes this proble" hard is the last re@uire"ent to tell in every casewhether the ball is light or heavy. #here are "any ways to find the oddBball! but

& only now one way to really do this proble"! and it is elegant because inevery case but one! the scale could balance or go either way on each weighing.#his is "y way4

eighing 1. Put four on each side of scale. &f they balance! it is case . &f not!then ,.

eighing 2! ase . Put three of the four unnown DunweighedE balls on oneside of scale and of the < nor"al fro" first weighing on other. &f they balanceit is case a! if not b.

eighing ! ase a. Put the only unweighed ball on balance with any otherDnor"alE ball. &f it goes up! it is light! if down it is heavy. &f it balancesso"ething is wrong+ D#his is the one case where it cannot balanceE.

eighing ! ase b. Gou now now that the odd ball is a"ong the three newones (ust weighed! and you already now whether it is light or heavyDdepending on if that side of scale went up or downE. eight two of those threeagainst each other. &f it balances! the other one is the odd ball. &f not! the onethat goes in the direction you already now is the odd ball. #hat is! if you

already new one of the three was light! then the side that went up has the light ball.

eighing 2! ase ,. #ae the four balls fro" the side of the scale which wentdown and put two on each side of the scale. lso tae two of the four fro" theside that went up and put on each side of the scale! noting which are which.


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Now there are three balls on each side of the scale. &f it balances! it is case ,a!if not ,b.

eighing ! ase ,a. Now we now that one of the two balls used in the firstweighing but not the second is light. eight either one against any other

standard ball. &f it goes up it is light! otherwise the other one is.

eighing ! ase ,b. hichever side went down has two balls that "ight beheavy! or there could be one light one on the other side. eight the two

possible heavies against each other. &f one side goes down! that is the heavy ball. &f they balance! then the one the "ight be light on the other side is indeedlight.

18. prisoner was given a chance to be blindfolded and pic one ball fro" two bowls that would contain a total of 60 white and 60 blac balls. hoosing white

"eant freedo"! blac "eant death. e ased if he could divide the balls between the bowls before he was blindfolded and his re@uest was granted.hat is the best way to divide the balls between the bowls/

e ased that all the balls be put in one bowl e$cept one white ball in the other.#here was a 1I2 chance of getting the bowl with the white ball! and 100chance of getting a white ball in that case. 5ven if he got the other one! he stillhad a 3>I>> chance of life! which is nearly 1I2. #hus the total odds are about1I2 Dif right bowlE K 1I3 Dhalf the ti"e if wrong bowlE L I3.

17.ow can three "issionaries and three cannibals cross a river in a canoe thatholds at "ost two people! if the cannibals "ust never outnu"ber the"issionaries on either side/

#his answer is being added to this page on 17 Feb 2007. ,efore today! & had anadditional constraint in the proble" that the two cannibals could not crosstogether. For the answer! & said that it was hard and &*d have to co"e bac to it.#oday & ca"e bac to it! and it was not only hard! it is easy to show it isi"possible! and no one has told "e after several years+ #he only possible first"ove would be that one "issionary and one cannibal cross together! and then

the only "ove would be that the "issionary brings the canoe bac. #hen thereis no possible second "ove+ &f you saw that and didn*t tell "e! sha"e on you.-o today & dropped that "istaen constraint and here*s an answer4

-end two cannibals over first and have one bring bac the canoe. #hen repeatthat! and then send two "issionaries across and have one bring bac a cannibal.#hen send both the re"aining "issionaries over! and have a cannibal return the


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canoe. Now all the cannibals are on the near side of the river! and the"issionaries on the far side. Now (ust have the cannibals bring the rest over intwo trips and you*re done.

1.#he following puzzle is about two actual historial people.

9eorge and 5velyn never "et but they carried on writing until late in life. &t has

been said that 5velyn loved 9eorge! but she was too old for hi". 9eorge"arried in 1


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is na"e is :ennis ,rent,riggs. &f you got :an ,rent,riggs! you did great! butyou fell into the trap ofaccepting a highly probable

short first na"e that was not part of the puzzle before

looing at all possible answers. &f you had found C:ennis!C there would have been no doubt in your "ind.

#he first letter of each line and of the first three words for" an acrostic!i"plying that his initial "ight be :,,. #he code is to count forward a fi$ednu"ber of letters to find a hidden na"e. For e$a"ple! begin on the C,C inC,earC and count forward every three letters. Gou find that it spells out C,rent.C

Note in the illustration that it for"s a perfect cross! designed to "ae it clearthat it is not there by chance! in order to help you discover the code.

Now count every 6 letters beginning at the first letter to get C:,rent,riggs.C#hat confir"s the suspicion of the first initials! but leaves the first na"e

unnown. -tarting with the first C:C you can find the na"e :an by countingevery seventh letter. ,ut to be thorough you should try all possible counts.ounting every 11th letter yields C:ennis.C ,ecause that is a "uch longer na"eDand hence i"probable to be there by chanceE! and because it begins and endson the first and last letter! it should be clear that it is the one the puzzle designer had in "ind. Note that even the capitalization of the letters co"es out correctly.

The verse in rows separated y three letters!

The verse in rows separated y "ive letters!


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,y the way! these codes are the sa"e as used in the very controversial soBcalledC,ible odes!C e$cept that those codes can run forward or bacward.

My Favorite Math Puzzles

by John P. Prattlast updated 7 Nov 200<

None of the following puzzles has a tric answer! or unwarranted assu"ptions. #heyare arranged appro$i"ately in order of difficulty% the first few can be done by

ele"entary school children! but they "ay need to be taught how to thin. & tried toinclude a wide variety of types of puzzles! so that each would teach a new lesson infinding creative solutions by breaing a variety of inds of "ental blocs. 'oreover!"ost of the" can be solved "entally by using a (udicious point of view.

1. #hree "en each paid M10 to share a M0 hotel roo". )ater! the "anager felt badabout overcharging the" for a M26 dollar roo"! so he gave the bellboy 6 oneBdollar bills to return to the". #he bellboy returned one dollar to each of the"en! but ept two for hi"self. -o the "en paid M27 for the roo"! plus M2 to the

bellboy for a total of M2>. here did the e$tra dollar go/

2. &f you had fifty co""on ?.-. coins that added up to e$actly a dollar! how "anydi"es would you have/

. race car is going around a trac that is a s@uare! one "ile on a side. &f it goes80 "ph around the first two sides and 0 "ph on the the third side! how fast"ust it go to average 80 "ph all the way around/

3. &f a hier averages 2 "ilesIhr uphill and 8 "ilesIhr co"ing bac the sa"e trail!what is his average speed going both ways/

6. #he coach! who also taught "ath! said he*d give each of the eleven boys on the

football tea" the sa"e a"ount of "oney to buy candy! provided that each spentthe "oney differently on the 2! ! 3! or 8 cent candy. &f he gave each thes"allest a"ount with which that could be done! how "uch did each receive/

8. &f 8 anteaters can eat 8 ants in 8 "inutes! how "any anteaters would it tae toeat 100 ants in 100 "inutes/ nother variation is4 &f a chicen and I3 can layan egg and I3 in a day and I3! how "any chicens does it tae to lay a dozeneggs lay in a wee/ DJr. ighE.


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7. Four blac cows and three brown cows give as "uch "il in five days as three blac cows and five brown cows give in four days. hich color cow gives the"ost "il/


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16. spoonful is re"oved fro" a cup of wine and placed in a cup of water andstirred well. #hen a spoonful is re"oved fro" the resulting solution and put

bac into the cup of wine. &s there "ore water in the wine or vice versa/

18. spider is one foot fro" the floor in the "iddle of 12 foot s@uare end wall of a

0Bfoot long roo". fly is situated in the "iddle of the other end wall! 1 footfro" the ceiling. hat is the shortest waling distance fro" the spider to thefly/

17. boo has 800 pages and an average of 1 typographical error per page. hatis the probability of finding n of the" on any one page Dfor n L 0! 1! 2! or anyintegerE. Dint4 use a Poisson distribution.E

Math Puzzle Answers by John P. Pratt

last updated 1 'ar 2007

1. #hree "en each paid M10 to share a M0 hotel roo". )ater! the "anager felt badabout overcharging the" for a M26 dollar roo"! so he gave the bellboy 6 oneBdollar bills to return to the". #he bellboy returned one dollar to each of the"en! but ept two for hi"self. -o the "en paid M27 for the roo"! plus M2 to the

bellboy for a total of M2>. here did the e$tra dollar go/

#his is sort of a tric @uestion! but it usually taes people a surprisingly longti"e to get it. #he tric is the word CplusC. &n reality! the M2 should besubtracted fro" Dnot added toE the M27 to get M26 paid for the roo". #here is no"issing dollar.

2. &f you had fifty co""on ?.-. coins that added up to e$actly a dollar! how "anydi"es would you have/

Gou would have 2 di"es. Gou "ight also have 30 pennies and < nicels! or you

"ight have 36 pennies! a @uarter and 2 nicels.

. race car is going around a trac that is a s@uare! one "ile on a side. &f it goes80 "ph around the first two sides and 0 "ph on the the third side! how fast"ust it go to average 80 "ph all the way around/


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No! the answer is not >0 "ph! and it is also not 120 "ph. #he average ti"e isthe total distance divided by the total ti"e. 80 "ph is 1 "ile per "inute. #oaverage one "ile per "inute around a 3 "ile trac! it "ust finish the race infour "inutes. &t spent one "inute on the first leg! another on the second! and 2"inutes on the third. #hus it has used up all four of the "inutes needed+ #hus!

no "atter how fast it goes! it cannot average 80 "ph all the way around.

3. &f a hier averages 2 "ilesIhr uphill and 8 "ilesIhr co"ing bac the sa"e trail!what is his average speed going both ways/

#his is si"ilar to the last proble"! in that you cannot si"ply average speedsand get 3 "ph as the correct answer. #he answer is easiest to see if you picany distance at all for the length of the trail! say 8 "iles. 9oing up taes hours and co"ing down taes one hour. #hat is 12 "iles in 3 hours! so theaverage is "ph.

6. #he coach! who also taught "ath! said he*d give each of the eleven boys on thefootball tea" the sa"e a"ount of "oney to buy candy! provided that each spentthe "oney differently on the 2! ! 3! or 8 cent candy. &f he gave each thes"allest a"ount with which that could be done! how "uch did each receive/

#,

8. &f 8 anteaters can eat 8 ants in 8 "inutes! how "any anteaters would it tae toeat 100 ants in 100 "inutes/ nother variation is4 &f a chicen and I3 can lay

an egg and I3 in a day and I3! how "any chicens does it tae to lay a dozeneggs lay in a wee/ DJr. ighE.

&f 8 can eat 8 in 8 "inutes! then those sa"e 8 can eat 12 in 12 "inutes! 23 in 23"inutes or 100 in 100 "inutes. Gou can thin of it as all 8 as woring on oneant in one "inute and then going to the ne$t ant.s for the chicens and eggs! you can co"e up with a for"ula! which & did in"y last posted solution! but it gives no insight into how to do the proble" orwhat are the principles at wor. =ne "ethod & use to solve proble"s is to beginwith so"ething & now! which is whole nu"bers of chicens. -uppose 2

chicens could lay 2 eggs in 2 days. )et*s try varying so"e of those nu"bers tosee what happens. &f 2 chicens lay 2 eggs in 2 days! then it taes each chicentwo days to lay an egg. -o 1 chicen would lay 1 egg in 2 days! and 37chicens would lay 37 eggs in 2 days. -o we learn Rule #$ We can %ultiplythe "irst two nu%ers y the sa%e nu%er and &eep the third the

sa%e! Now let*s eep the "iddle nu"ber fi$ed. ow long would it tae for 1chicen to lay 2 eggs/ alf as "any chicens woring would tae twice as long


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to produce the sa"e output. -o 1 chicen could lay 2 eggs in 3 days! and 3chicens could produce 2 eggs in 1 day. -o we have Rule '$ (" we %ultiply thenu%er o" wor&ers y N and &eep the output "i)ed* then we %ust divide

the ti%e y N! Now suppose we eep the nu"ber of chicens fi$ed. ow longwould it tae 2 chicens to produce 3 eggs/ &t would tae 3 days. nd 2

chicens could produce 8 eggs in 8 days. -o we have Rule +$ We can &eep the"irst nu%er "i)ed and %ultiply the last two y the any nu%er and it will

e true! Now let*s solve the proble" using these rules.'ethod 14 ?se two of the rules to change the nu"bers to the desired ones. &f7I3 chicens can lay 7I3 eggs in 7I3 days! how "any chicens would it tae tolay 12 eggs in 7 days/ e can use either Aule 2 or Aule to change the nu"ber of days to 7. ?sing rule ! we "ultiply the last two nu"bers by 3 and get that7I3 chicens can produce 7 eggs in 7 days. Now use Aule 1 to change thenu"ber of eggs while leaving the days unchanged. e "ultiply the first twonu"bers by 12I7 Dto get a dozen eggsE and get the answer that chicens canlay a dozen eggs in a wee. D5$ercise for student4 9et sa"e result by usingAule 2.E'ethod 24 #his proble" was sub"itted by "y friend 'oray ing who prefersthe "ethod of always beginning by applying Aule 1 first to find the output ofone worer. 'ultiplying the first two nu"bers by 3I7 "eans that 1 chicen can

produce 1 egg in 7I3 days. Now we can apply Aule and "ultiply by 3 to getone wee4 1 chicen can produce 3 eggs in 7 days. nd now Aule 1 again and"ultiply by three to get the desired nu"ber of eggs4 chicens can lay 12 eggsin 7 days. #hat too an e$tra step! but it is easier to re"e"ber and conceptually

very easy. hich of the Aules would have solved the anteater proble" in onestep/

7. Four blac cows and three brown cows give as "uch "il in five days as three blac cows and five brown cows give in four days. hich color cow gives the"ost "il/

& don*t see a way to solve this by inspection because the nu"bers are too close.=ne way to solve it is by beginning with "y Aule 2 Dwhich eeps productionthe sa"eE fro" the last proble". #hus! to see how "any of the latter "i$ture

would be needed to produce the sa"e "il in 6 days D6I3 the ti"eE! we "ultiplythe cows by 3I6. #hat "eans that 2.3 blac cows and 3 brown cows produce thesa"e "il in 6 days as do four blac and three brown. #hat "eans that 1 browncow is producing as "uch as 1.8 blac cows! so the brown is "ore.

Now let*s solve it with algebra. #he rate of production of each cow ti"es theti"e e@uals the a"ount of "il. )et r , be the rate for ,lac cows and r b be for

brown cows. #hen D3Or , K Or bEO 6 days L DOr , K 6Or bEO 3 days. -olving this


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e@uation for the ratio of rates r ,Ir b yields 6I

. ithout writing down any trial solutions! prove that the following digitsubstitution proble" has no solution4

E L E V E N

- T H R E E

E I G H T

#,

10. tourna"ent had and , divisions! each with fro" to 10 contestants. 5achdivision was a roundBrobin! in which every contestant plays every othercontestant. =ne contestant was not allowed to register late because there werealready so "any "atches to be played. owever! that decision was reversedwhen he showed that allowing hi" to enter would actually decrease the totalnu"ber of "atches by 2! provided that his friend were allowed to changedivisions. ow "any contestants ended up in each division/

#,

11.#wo professors "et again after several years and one said he was now "arriedwith three children. hen the other ased their ages! the first said that the

product of their ages is 8. #he second ased for another clue and the firstased if he could see the nu"ber on the house across the street. hen the othersaid yes! the first said that the su" of their ages e@ualed that nu"ber. #hesecond said he still could not deter"ine their ages. #hen the first said that hisoldest child has red hair. Finally the second new their ages. hat were their

ages and how did he now/

#his is "y all ti"e favorite proble". #o do it! first write down all the real possibilities that the nu"ber on that building "ight have been. ssu"inginteger ages you get the following which e@ual 8 when "ultiplied4

ges L 1!1!8 Dsu" L


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ges L 1!2!1< Dsu" L 21Eges L 1!!12 Dsu" L 18Eges L 1!3!> Dsu" L 13Eges L 1!8!8 Dsu" L 1Eges L 2!2!> Dsu" L 1E

ges L 2!!8 Dsu" L 11Eges L !!3 Dsu" L 10E

#he big clue is that the second professor :&: N=# N= after having beentold the su" e@ualled the nu"ber on the house. hy didn*t he now/ #he onlyreason would be that the nu"ber was thirteen! in which case there are two

possible answers. For any other nu"ber! the answer is uni@ue and the professorwould have nown after the second clue. -o he ased for a third clue. #he cluethat the oldest had red hair is really (ust saying that there is an ColdestC!"eaning that the older two are not twins. ence! the answer is that the redheadis > years old! and the younger two are both two years old.

12. boy spent e$actly one dollar and bought e$actly 100 pieces of candy!including so"e at M .06! M.02! and 10 for M.01. ow "any pieces of each indof candy did he buy/

11 pieces of the nicel candy! 1> of the 2 cent! and 70 of the 10Ipenny.

1.=n a #; @uiz show a contestant is shown three closed doors and told that twoof the" have nothing behind the"! but that one has a new car as a prize behind

it. #he contestant "aes her choice of doors where she thins the prize is. #henone of the other two doors is opened where there is no prize! and the contestantis ased if she would lie to change her guess. :o the odds favor changing theguess/ hy or why not/

#he odds favor changing your answer. #his proble" is fa"ous because 'arilynvos -avant! the Cs"artest person in the world!C answered this @uestion in hercolu"n in Parade! in one sentence lie & (ust did. #hen several fa"ous "ath

professors challenged her! saying that it "ade no difference! that the chance beca"e 1I2 for each after the other door was opened. 'arilyn turned out to beright and gave a nice! short proof. &*ll give a different proof which "y "otherused.

hen & ased "y nonB"ath"atical ninetyByearBold 'other this puzzle! & wasastounded that she got the right answer in seconds! reasoning as follows. -hesaid! Cthe chance is 1 in that it is behind the door she chose! and zero chancethat it is behind the one that was opened! so there "ust be a 2 in chance that it


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is behind the other one.C #hat is e$actly right. #o see why! pretend that therewere 100 doors instead of three. -uppose they ept opening all >< other doors!so that only the original door chosen was left and one other. Now would youchange doors/ #here is only a 1I100 chance that the first door was right! and ahuge chance D>>I100E that the other is right.

13.&f #o" is twice as old as oward will be when Jac is a old as #o" is now! whois the oldest! ne$t oldest and youngest/

#,

16. spoonful is re"oved fro" a cup of wine and placed in a cup of water andstirred well. #hen a spoonful is re"oved fro" the resulting solution and put

bac into the cup of wine. &s there "ore water in the wine or vice versa/

#here are e@ual a"ounts in each. &*ll co"e bac and show why so"e day.

18. spider is one foot fro" the floor in the "iddle of 12 foot s@uare end wall of a0Bfoot long roo". fly is situated in the "iddle of the other end wall! 1 footfro" the ceiling. hat is the shortest waling distance fro" the spider to thefly/

No the answer is not 32 Dgoing straight up or down and aroundE. #he only way &now to do this proble" is to drawn pictures of every possible way to cut theroo" apart and have the spider wal in a straight line between the points. Gou

have to now the Pythagorean for"ula that su" of the s@uares of the legs of aright triangle e@uals the s@uare of the hypotenuse. Gou*ll now when you havethe answer because it co"es out in round nu"bers.

17. boo has 800 pages and an average of 1 typographical error per page. hatis the probability of finding n of the" on any one page Dfor n L 0! 1! 2! or anyintegerE. Dint4 use a Poisson distribution.E

Gou need higher "ath for this one! but & put it in because of how nifty theanswer is. &t turns out that the probability of n errors per page is 1IDen+E where e

L 2.71


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Jacob's Missing Descendant

Jacob's extended family at the time he moved to Egypt is listed inthe Bible, but some informationabout one descendant may have

been purposely hidden. If there is no mistake in the following

summary and interpretation of the Biblical account, what canyou

logically deduce about the identity of Jacob's missing descendant

1. ll seventy living souls of the house of Jacob! including all of his living "ale andfe"aledescendants! were in 5gypt when he arrived there with those who acco"paniedhi". D9en. 3848! 27E.

2. -i$tyBsi$ of Jacob*s descendants ca"e to 5gypt with hi". #his count includes onlyJacob*s literaloffspring% none of his sons* wives is included D9en. 38428E.

. 5$cept for Joseph and his two sons! who already resided in 5gypt D9en. 38427E!Jacob too withhi" all of his sonDsE! his sons* sonDsE! his daughterDsE! his sons*daughterDsE D9en. 3847E! and all ofhis greatBgrandchildren DClittle ones!C 9en. 3846E.

3. #hese are the na"es of Jacob*s descendants when they had all arrived in 5gypt!along with subtotals for each of his four wives D)eah! ilpah! Aachel! and ,ilhahE4

a. !eah had "" living descendants. #er sons were $euben, %imeon,

!evi, Judah, Issachar,and &ebulun, and her daughter was inah.

$euben's sons were #anoch, (hallu, #e)ron and*armi. %imeon's

sons were Jemuel, Jamin, +had, Jachin, &ohar, and %haul the son of

a*annaanitish woman. !evi's sons were ershon, -ohath and erari.

Judah's sons were Er,+nan, %helah, (hare), and &erah, but Er and

+nan had died previously. (hare)' sons were#e)ron and #amul.

Issachar's sons were /ola, (huvah, Job and %himron. &ebulun's

sonswere %ered, Elon and Jahleel 0en. 12345678.

b. &ilpah had 62 living descendants. #er sons were ad and 9sher.

ad's sons were &iphion,#aggi, %huni, E)bon, Eri, 9rodi, and 9reli.

9sher's sons were Jimnah, Ishuah, Isui andBeriah, and %erah was

their sister. Beriah's sons were #eber and alchiel 0en. 123625648.


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c. $achel had 61 living descendants. #er sons were Joseph and

Ben:amin. In Egypt, the sonsof Joseph and his wife 9senath,

daughter of an Egyptian priest, were anasseh and

Ephraim.Ben:amin's sons were Belah, Becher, 9shbel, era,

;aaman, Ehi, $osh, uppim, #uppim,and 9rd 0en. 1236 living descendants. #er sons were an and ;aphtali.

an's son was #ushim.;aphtali's sons were Jah)eel, uni, Je)er and

%hillem 0en. 123="5=78.

my favorite logic puzzles

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