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Shakuntala Devi Puzzle book: Puzzles to Puzzle perpemethico.ml 3. George Summers Puzzle book:Download George J. Summers - Test Your Logic (50 Puzzles) -. PDF processed with CutePDF evaluation edition perpemethico.ml THE GREAT BOOK OF PUZZLES & TEASERS GEORGE J. SUMMERS. Documents Similar To The Great Book Of Puzzles And Teasers (gnv64).pdf. Puzzles and Teasers by George j Summers PDF. Uploaded by. latika · Fantastic.

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Puzzles and Teasers by George j Summers PDF - Download as PDF File .pdf), Text File .txt) or read online. Tata. TEST YOUR LOGI!qM Jno] Ol S! l! A:q 9 The Round Anthony, Bernard, and Charles played a round of card games, each game having exactly. This book consists of 75 utterly original and totally tantalizing brain teasers from master puzzler George J. Summers. He brings out fascinating challenges in.

George J. A:q 9 The Round Anthony, Bernard, and Charles played a round of card games, each game having exactly one winner. The player who first won three games was to be the winner of the round. No player won two games in succession. Anthony was the first dealer, but not the last. Bernard was the second dealer. The players sat in fixed positions around a table with the player on the current dealer's left the next dealer.

Lancer sit? Notice that if the" head" of the top factor falls to the rear it becomes the product. Which one of the ten digits does M represent? Dora dealt one card to Lois, then one card to Rose, then one card to herself, and repeated this order until all the cards were dealt.

After the pairs in each hand were removed, at least one card remained in each hand; the number of cards in the three hands totaled 9. Of the remaining cards, Lois' and Dora's hands together formed the most pairs, and Rose's and Dora's hands together formed the least pairs. Who was dealt the singleton? OMl , q sJ! Two women, each of whose coins totaled 60 cents, had the same number of silver coins; no denomination of coin was held by both women.

Two women, each of whose coins totaled 75 cents, had the same number of silver coins; no denomination of coin was held by both women. Agnes' check was for 10 cents, Betsy's check was for 20 cents, Cindy's check was for 45 cents, and Delia's check was for 55 cents. Each woman paid the exact amount of her check. Two women, who were sisters, had the same number of coins left after paying their checks. Which two women were sisters? In each tournament just four sets were played, as follows: The winner of only one set was the same in both tournaments.

Alan won the first tournament. In each tournament, only the winner of the tournament did not lose a set. Who won the second tournament? A tie is not possible in a set. W lSU! Which one of the ten digits is absent? On only one day of the week are all three interns on call.

No intern is on call on three consecutive days. No two interns are off on the same day more than once a week. The first intern is off on Sunday. The second intern is off on Thursday and Saturday. The third intern is off on Sunday. Which day of the week are all three interns on call?

Their occupations were as follows: If the lead and the director were related, the director was a singer. If the lead and the director were not related, the director was a man. If the lead and the director had the same occupation, the director was a woman.

If the lead and the director had different occupations, the director was a White. If the lead and the director were the same sex, the director was a dancer. If the lead and the director were different sexes, the director was a Black. Who was the lead? One is a pianist, a second is a violinist, a third is a flutist, and a fourth is a drummer.

On a day they were seated around a square table: The person who sat across from Burton was the pianist. The person who sat across from Donald was not the flutist. The person who sat on Arlene's left was the violinist. The person who sat on Cheryl's left was not the drummer. The flutist and the drummer were married. Who is the drummer? One man's apartment was in the middle, adjacent to each of the other two men's apartments.

Each man owned only one pet: Austin lived next to a cigar-smoker. Brooks lived next to a dog-owner. Calvin lived next to a tea-drinker.

No pipe-smoker was a tea-drinker. At least one cat-owner was a pipe-smoker. At least one coffee-drinker lived next to a dog-owner. No two designations such as cigar-smoker, dog-owner, tea-drinker, etc. Whose apartment was in the middle? The number of blocks along the border of each town is a whole number; each block through each town is parallel to one pair of blocks along the border.

Arlington has the least number of blocks along its northern border, Burmingham has three more blocks along its northern border than Arlington has, and Cantonville has three more blocks along its northern border than Burmingham has. The number of interior blocks through each of two of the towns equals the number of blocks along its entire border. In which one of the three towns does the number of interior blocks through it not equal the number of blocks along its entire border?

A "block" refers to a length, not to an area. However, two of the dice above are alike and one is different in respect to the orien-tation of the faces. If you find it difficult to visualize a die's six faces, you might draw a multiview die as shown below. The bottom face will be the only face not seen. S oMlluql os saqn: The four men, each of whose coins totaled one dollar, had the same number of silver coins.

Amos had exactly three quarters, Bert had exactly two quarters, Oem had exactly one quarter, and Dirk had no quarters. The four men had to pay the same amount; three paid the exact amount, but the fourth required change.

Who required change? Rd aq lOUUR:: W1alap uaql q:: WJalaa 27 The Doctor Mr. Blank has a wife and a daughter; the daughter has a husband and a son.

The following facts refer to the people mentioned: One of the five people is a doctor and one of the other four is the doctor's patient. The doctor's offspring and the patient's older parent are of the same sex. The doctor's offspring is a.

Who is the doctor? I SJjuq: Astor, Mr. Brice, and Mr.

Crane were seated around a table as shown below. Crane Mrs. Bri", t Astor Mrs. Crane At the table: Exactly three people sat next to at least one murderer. Exactly four people sat next to at least one extortionist. Exactly five people sat next to at least one swindler. Six people sat next to at least one thief. Of the types of crime: No two types of crime were committed by more than one person. One person committed more types of crime than each of the other persons. Of the people: Astor each committed exactly one type of crime, though not the same type.

Brice were both swindlers. Crane were both thieves. More women than men were swindlers. Determine separately the possible seating arrangements of the murderers, of the extortionists, of the swindlers, and of the thieves. Then determine the number of persons who committed four types of crime, just three types of crime, just two types of crime, and just one type of crime. Finally determine the specific types of crime committed by each person. J"a u"'ll aul!.

J fo GadAI a.

This situation contradicts [3]. So Adrian orders only pork. Then, from [2], Carter orders only ham. So only Buford could hIlve ordered ham yesterday, pork today. If Val's father were Chris, then Chris' sibling would have to be Lynn. Then, from [2], Lynn's daughter would have to be Val.

This relationship is incestuous and is not allowed. So Val's father is Lynn. Then, from [2], Chris' Sibling is Val. So Lynn's daughter is Chris. Then, from [1], Val is Lynn's son. Therefore, Chris is the only female.

So, from [2], the division of the number of nurses according to sex must be such that the number of males is less than six. From [3], the number of female nurses must be less than the number of male nurses. So there must be more than four male nurses. So there must be no more than nine nurses, consisting of five males and four females, and there must be no less than six male doctors. Then there must be only one female doctor to bring the total to If a male doctor is not included, [2] is contradicted.

If a male nurse is not included, [3] is contradicted. If a female doctor is not included, [4] is contradicted. If a female nurse is not included, no fact is contradicted; so the speaker isfemale and is a nurse. From [7], Freeman will not marry Ada or Cyd. From [2], [5], and [6], either Cyd or Deb must have the same occupation as Bea and Eve; so Bea and Eve are secretaries.

From [7], Freeman will not marry Bea or Eve. By elimination, Freeman will marry Deb, who must be over 30 and a teacher. The remaining characteristics of the other four women can be determined from previous reasoning. Eve must be under 30 and Bea must be over Cyd must be a secretary and Ada must be a teacher.

G, H, and I represent different digits; so one number was carried from the right-hand column to the middle column, and a different number was carried from the middle column to the left-hand column. The only column sum, less than or equal to 27 for which this is true is It follows that FGHI equals What triplets of different digits result in a sum of 19, with no digit in the triplet equal to zero, 1, 2, or 9?

So A represents 8.

The two possible additions are: From [5], if Annette is rich she is artistic. From [1] and [2], if Annette is neither rich nor intelligent she is artistic. So, in any case, Annette is artistic. From [4], if Claudia is beautiful she is artistic. From [5], if Claudia is rich she is artistic. From [1] and [2], if Claudia is neither rich nor beautiful she is artistic. So, in any case, Claudia is artistic. Then, from [1], Bernice is not artistic. Then, from [4], Bernice is not beautiful.

So, from [1] and [2], Bernice is intelligent and rich. Then, from [1], Annette and Claudia are both beautiful. Then, from [2] and [3], Annette is not intelligent; so, from [1], Claudia is intelligent.

Then, from [1] and [2], Annette is rich and Claudia is not rich. So Alice is the tennis player. If just five games were played, the winner of the round would have won the first, third, and fifth games, from [2]. But from [3], [4], and [5], each man would have been the dealer for one of these games. This situation contradicts [6], so exactly six games were played. Since exactly six games were played, Charles was the dealer for the last or sixth game, from [3], [4], and [5]. From [1], the winner of the last or sixth game won the round; so, from [6], Anthony or Bernard won the last or sixth game and, thus, the round.

If Anthony won the sixth game he could not have won the first game or the fourth game, from [6]; nor could he have won the fifth game, from [2]. Then he could only have won the second and third games, a situation which contradicts [2].

So Anthony did not win the sixth game. Then Bernard must have won the sixth game and, thus, Bernard won the round.

There are four possible sequences of wins as shown below A represents Anthony, B represents Bernard, and C represents Charles. S9 So there are six possibilities for the values of A, B, C, and D; the possi-bilities are shown in the chart below.

CB D A a 37 1 3 b 37 2 6 c 37 3 9 d 74 2 3 e 74 4 6 f 74 6 9 Since each letter represents a different digit, possibilities a , c , and e are immediately eliminated. DOing the actual multiplication in b , d , and f to determine E, F, and G in each case, one gets: So D represents 2.

If both [2] and [4] are true, then Curtis killed Dwight and, from [I], statements [5] and [6] are both false.

But if Curtis killed Dwight, [5] and [6] cannot both be false. So Curtis did not kill Dwight. Then, from [II], it is impossible for just one of statements [1], [3], and [5] to be true, as required by [I]. So [1], [3], and [5] are all false and [6] is the other true statement. Since [6] is true, a lawyer killed Dwight. Since Curtis did not kill Dwight from previous reasoning, Barney is not a lawyer because [3] is false, and Albert is a lawyer because [1] is false it follows that [4] is true, [2] is false, and Albert killed Dwight.

However, the spots for two, three, and six can have either of the following orientations: If die B were like die A, the two on die B would have an orientation opposite to that shown. So die A and die B are not alike. If die C were like die A, the three on die C would have an orientation opposite to that shown. So die A and die C are not alike. If die C were like die B, the six on die C would have an orientation the same as that shown.

So die A is different. Assumption [1] cannot be applicable because Beth's statement cannot be true under its application. So assumption [2] is applicable. Vera's dressing room borders on Adam's dress- ing room and on Babe's dressing room. Solution Scheme, page 68; Solution, page Statement B: Statement A and statement B are not both lies.

Gregory made one of the statements, his father made another of the statements, and his son made the remaining statement. Each father and son mentioned in the statements refers to one of the three men. Each man either always tells the truth or always lies. Orientation and Solution Scheme, page 70; Solution, page Statement A Statement B Statement C Write "true" or "false" for each statement so that no condition is contradicted and so that the truth or falseness of each man's statement is not contradicted.

Who paddled twice? Orientation and Solution Scheme, page 74; Solution, page He is married. To find a woman whose name satisfied the conditions above in relation to EDWIN would have been impossible. Has no third letter of the alphabet because 3 is hi unlucky number. Has its letters in alphabetical order when the first letter and the second letter are interchanged. Orientation and Solution Scheme, page 75; Solution, page P, Q, R, and S wish to cross a lake in a canoe that holds only two persons.

This puzzle involves a wledge of women's first names. Arguments One of four people-two men, Aubrey and Burtoll and two women, Carrie and Denise-was murdered The following facts refer to the people mentioned [1] Aubrey's sister argued exactly once with Carri I legal husband after the murder.

Who was the victim? The game requires: That the player who has to take the last chip loses. That Amelia now have her turn. Orientation and Solution Scheme, page 79; Solution, page If Amelia doesn't make the one right each time it is her tum, Beulah gets to be in the winning tion. Possible Possible Possible game game game o 00 o 00 0 00 -- -- -- - ow the chips remaining after each allowed draw.

Cross off any unnecessary "pile" 79 I] Each letter represents a different digit. No letter represents zero. Orientation and Solution Scheme, page 83; Solution, page Certain orders are equivalent: Who got the most correct answers? Orientation and Solution Scheme, page 86; Solution, page Sitting Ducks Mr. Astor, Mr. Blake, Mr.

Crane, and Mr. Davis were seated around a table as shown at right. At the table: Astor was insulted by Mr. Blake who sat next to her on her left. Blake was insulted by Mrs. Crane who sat opposite him across the center of the table. Crane was insulted by the hostess who was the only person to sit next to each one of a married couple.

Orientation and Solution Scheme, page 87; Solution, page Consideration of the total number of correll answers also eliminates much trial and error. The High Suit Wilson, Xavier, Yoeman, and Zenger were playing a card game in which three cards from each player's holding remained to be played and in which one of four suits-clubs, diamonds, hearts, spades-was the high suit.

The play of four cards, one from each player's holding, was a trick; the suit of the card played first in a trick was the suit led.

Wilson's holding - club heart diamond Xavier's holding - clu,b spade spade Yoeman's holding- club heart heart Zenger's holding - spade diamond diamond [2a] A player had to playa card in the suit led, if possible, at each trick. Orientation and Solution Scheme, page 90; Solution, page Remaining tricks Pirst Second Third Wilson's holding Xavier's holding Yoeman's holding Write "c" club , "D" diamond , "H" heart , or "S" spad in each box so that no condition is contradicted.

The Fight From [I], Charles is not the older fighter. From [3]. Bernard is not the younger fighter. So either: Case I. Anthony is the older fighter and Charles is th younger fighter. Case II. Bernard is the older fighter and Anthony is th younger fighter. Case Ill. Bernard is the older fighter and Charles is th younger fighter. From [3], Charles is the taller fighter for Case I and Charle. From [2], Anthony is not the shorter fighter; so Case I i eliminated.

Then Case II is the correct one and Charles is not fighting. Then, from [2], Bernard is younger than Charles so Chari is the oldest of the three and Anthony is the youngest and. If you were told anyone of the letters in then you would not be able to determine whether the ber of vowels in my secret word is one or two. So none of letters in MOD is in my secret word.

So my secret is TIE. Then, from [2]: For ways la and lIa Brother is the worst player's sibling. Zita is the worst player. Then, from [3]: For ways la and lIa Brother is the best player. For ways Ib and lib Daughter is the worst player' sibling.

So, in any case, Zita's brother ;s the best player. After-Dinner Drink From [1] and [2] there are six possibilities: So you know Abigail always 0 ders the same drink tea after dinner. Then, from [1] and [2], 9 must go with nd. One digit must be 8. One digit must be 7. Then, from 1and [2], 7 must go with either 1 and 5 or 2 and 4. One 't must be 6. Then, from [1] and [2], 6 must go with either nd 5 or 3 and 4, rom the diagram no digit may be used in more than two From this and the fact that 9 goes with 1 and 3: If 8 goes with 1 and 4, then 7 goes with 2 and 4; 6 goes with 2 and 5.

If 8 goes with 2 and 3, then 6 goes with 2 and 5'. Case I is correct and, from the diagram, E must be 4. If Lee's spouse is Terry, then Terry's sister-in-law cannot Then, in any case, all three of Lee, Dale. So, from [2], Lee's spouse and Dale's sibling are both male. In summary: So Case I is the correct one and you know Dale is a married man.

Lee is a married woman, Dal- and Terry are brothers, and Lee is Terry's sister-in-law. Then, from [3], the seventh mark must be placed in square 5 and, from [2], the seventh mark wins for both X and O.

So the sixth mark must have been placed in a line already containing two of the opponent's marks- in either square 7 0 1 square 9; otherwise, either X or 0 would have been placed in square 5. But, from [3], the sixth mark could not have been placed in square7because square 5 would have been the required location for the 0 in square 7. So the sixth mark was placed in 96 rc 9 and was X. Then, from [I], the seventh mark will be 0I.

From [1], no one can have more than two of the ; so, from [I] and [2J, Kay is distributed only twice ay in one of two ways: Then Case I is u VI" and Case II becomes: Then, from [I], Brent is named May.

Hockey Scores From [I] and [3]: The team that lost the greatest numbcl of games lost the two games it played there were three loser and the teams could not have each lost one game.

So th team that lost the greatest number of games did not score I and did not score 5 and 4 together. The highest total of tWII scores achieved by this team is greater than the total of II least 7 6 and at least I achieved by some other team. So thl team scored 5 and 3 for a total of 8. Then the 5 score lost to the 6 score and the 3 score lost II the 4 score.

So the I score and the 2 score go together and th 6 and the 1 were scored by the same team. Let the teams b X, Y, and Z temporarily; then in summary: Then the Cougar achieved the highest total of its two scores.

The Journey From [3]: No two of the L-shaped fields are the same kind of field; these can be assigned letters, such as east-E, west-W. Then 1 W field, 5 is an E field, and 12 is an S field. So, from [2], each E field rye field and the east L-shaped field is a rye field. From [3], Alma did not visit Edna first.

So Brand 1 age must be a one-digit number. So Che le i age must end in zero. Then Chester's age must be a two-di'll number. From [4] and the fact that only Brandon's age i one-digit number, Brandon is J d fterent age in years from ,h, other two. Because Chester's age ends in zero, Ambrose's age mil I end in zero. So Ambrose and Chester may.

Of th thirty-six possible ways to complete the diagram, here is on. So, from [2], if the is a woman, the lawyer is a man. Then, from [3], the is a man. Because a contradiction arises from assuming doctor is a woman, the doctor must be a man. Horton's son is the youngest of the four and is a blood ive of each of the other three.

So, from [1], the doctor is Mr. Horton's son. Then you know the occupation of Mr. Horton's mother. The doctor may be than the lawyer. Then you don't know the occupation of else. If both occur twice, then the identical cubes look like: If only.

But in this case duplicate faces cannot occur at the arrow. In this case the 0 occurs on the right cube at the arrow. So, from [2], the 0 occurs on the left cube at the arrow. The 0 occurs on the unmarked face in the last diagram.

The Guards From [I] and [2], Ida stands guard on only two days each week. If Bob stands guard on Wednesday, then Art stands on Friday. Then Cab stands guard on Thursday and Bob guard on Saturday. Then Art stands guard on Monday. Cab stands guard on Tuesday. If Cab stands guard on Wednesday, then Art tands on Thursday. Then Bob stands guard on Tuesday and. Then Cab stands on Saturday. Eye color Eye color Eye color of of of Amos Bert Clem hazel blue hazel blue blue hazel hazel blue blue hazel blue hazel [I], then, a fourth brother has grey eyes in each case.

Equal Products From [1] and [2]: No letter can be 0, 5, or 7. So the smallest possible product is 8 x 9 or 72 and th product is a multiple of But the product cannot be any 01 72 x 2, 72 x 3, etc. So the product is A possible arrangement of digits is: Tees and Eis From [I] and [2]: If Annette's statement is true, all three cannot members of the Tee family and Cynthia cannot be the only one of the three who is a member of the EI family.

So, it Annette's statement is true, either: Annette is the only on'. If Annette's statement is false, Annette cannot be only one of the three who is a member of the EI family Bernice cannot be the only one of the three who i a ber of the Tee family.

So, if Annette's statement is fal e,: Cynthia is the only one of the three who is a member the Tee family or all three are members of the EI family. Cynthia may be a member of either mily in Case I and Cynthia may be a member of either family 'ase II. So you know only the name of Bernice's family EI. From [I] and [2], Carl cannot have had exactly o singletons unless one of them was the joker.

So Alec had singleton king and the singleton queen. Then, from, [3], each of Bill and Carl cannot have had more an one singleton. If Carl had no singletons, then- from Alec's Iding and I] and [2]- he had two kings and two queens.

So Carl had one singleton. Then-from [I], [2], and. Then simplify. Because only one salesperson was wrong, from [4], neither the second nor the third salesperson was wrong; otherwise, more than one salesperson was wrong.

So only the first salesperson was wrong. One possible price list, from [2] and [3], is: In summary, the married couples are either: So Case 1 is the t one and, from [I], Cass is the widow. The first logician to answer "Yes" must ve been told a letter that occurred only once in the list.

Then second logician knew what the word was if told T or E new letter , or if told the second letter of PAD that only once in the list: So the third I gician knew the word was not OAR. So the third logician new what the word was if told 0 some new letter , but not know what the word was if told A. Speaking of Children From [1], Aaron has at least 3 children and a number 1.

From [2], Brian has at least 4 children and a number III children from this sequence: From [3], Clyde has at least 5 children and a number 01 children from this sequence: Then the total number of children is at least 12 and, from [41 at most If the total number of children is even, Aaroll , must have an odd number of children; if the total number 01 children is odd, Aaron must have an even number of childr II Trial and error reveals the following information.

The total cannot h ' 18, 20, 21, 22, 23, or 24 because then no number of children could be known for anybody, contradicting [4]. So the total is So Aaron must have 6 children Then Brian and Clyde together must have 13 children. Then Brian must have either 4 or 8 children.

If Brian has children, Clyde has 9 children; if Brian has 8 children, Clyd has 5 children. So the speaker is Aaron. Larchmont's Chair t X represent one sex, let Y represent the other sex, and an X in chair a. Conditions [1] and [2] be used alternately to determine the couples X-l and Y-I, and Y-2, etc. Case I I is the correct case. Larchmont sat in either chair i or chair j. Larchmont sat in chatr j. Then, from [ll and 2 , the possible sequence of wms IS folIows: Each succeeding case represents the other chpicc for the preceding case.

From [4], Cases I, II,. In either case, Cheryl was the only player to win more than tW I games. Meeting Day From [I] and [2], the two-day man went to the health club for the first time this month on the Ist or the 2nd and th three-day man went to the health club for the first time thi month on the Ist, 2nd, or 3rd. Then, from [3], the two-day man went on the ist and the three-day man went on the 3rd. Then-from [I], [2], and [3]-the seven-day man went to the health club for the first time this month on either the 5th I the 6th.

In summary, either A or B below is true. Lou went on Monday, the 1st, and every two day thereafter; Moe went on Wednesday, the 3rd, -and every three days thereafter; and Ned went on Friday, the 5th, and every seven days thereafter. So po slblltty I correct. Possibility B reveals that Lou, Mae, and Ned mel the 27th of this month. Aye won one game, and Mr Bee won one game; or II. Aye won two games and Mrs.

Aye won one game; or I, Mr. Bee won one game. If I is correct, then: From I] and 4], Mr. Bee beat rs Bee in the first game. Then, from 4], only Mr. Bee uld have lost to Mr. Aye or Mrs. So I IS not correct. So III is correct. If Mrs. Bee won the first game, then " beat Mr. Bee in that game, from [I]. But then, from [I] [4], no one could have played against Mr. Aye in the seC l1i game.

So Mr. Aye won the first game against Mrs. Aye, frnll [I]. Then, from [4J, Mr. Aye beat Mr. Bee in the sec1III I game. Bee beat Mr. Aye in the third ganl So only Mrs. Bee did not "hse 'Q game. R, L, and D. C, U, E. It is now easy to see that: So eithcl Adam or Dawn occupies dressing room Il. Then, from [3], one of the following sets of occupation must exist: So the set is the correct one and, from [2], Vera occupies ng room IV.

So Adam dressing room V. But- from [1], [2], and [3]- if statement A is false, ment B or statement C is true because a false statement A one father always tells the truth and one father always So statement C 'cannot be false and must be true.

But- from [1] and [2]-if statement B is true, ment A or statement C is false. Case I true false true Case II false true true f Case I were the right one, then- from statement A and [2]-the speakers of statements A and C would both be ; and- from statement B and from [2] and [3]- the of statements A and C would both be sons because a statement B implies both sons always tell the truth or both always lie.

This situation is impossible from [1] because Gregory is both a father and a son; so Case I is inated. Then, from statement A and So, from [I], Gregory made statement A.

Crossing the Lake From [I], [2], and [4], Agnes paddled on at least on return trip. The person who paddled twice did not paddle on two forward trips because, from [I], she would then have had III paddle on a return trip, contradicting [4].

So the person who paddled twice paddled on at least one return trip. In summary, Agnes and the person who paddled twice each paddled on at least Ol e return trip. So Becky, Cindy, and Delia each paddled on one forward trip, from [I] and [4]. Because Cindy was in the canOl on two forward trips, she must have paddled on a return trip So Cindy paddled twice.

Six paddling sequences are possible. The sequences are shown below. Division by 6 leaves 1. In alphabetical order the letters found so far are FNN and the only other letter her name may contain is A I. The W in Edwin is the twenty-third letter of the aphah t and can only be represented as 23 x 1. Suppose Aubrey's sister is Carrie. Then, from [I], Carn husband is Burton. Then, from [I] and [2], Burton's sister I Denise. Suppose Aubrey's sister is Denise.

Then, from [I], Carri' husband is either Aubrey or Burton. Carrie Burton Denise? Case lIa. Denise Aubrey Carrie? Case IIb. Denise Aubrey Denise?

Case IIc. Denise Burton Denise?

In Case I; Aubrey must be the victim. Then the victim' legal spouse can only be Denise. This situation is impossibl , so Case I is eliminated. In Case lIb, either Burton or Carrie must be the victim. Burton is the victim, then Burtoh can have no legal spou. If Carrie is the victim, then and Aubrey argued exactly once- from [I - and [2]; this situation is impossible.

If brey is the victim, then Aubrey can have no legal spouse; situation contradicts [2]. If Carrie is the victim, then ise and Burton argued exactly once- from [I]- and twice- [2]; this situation is impossible. So Case lIe is eliminated. Then Burton must be the. From [2], Burton's spouse can only be Denise.

From [I Amelia takes Amelia takes and [3]: Amelia loses Amelia loses From [1] Amelja takes Amelia takes and [3]: Amelia loses Amelia loses From [IJ and [3]: Amelia takes one chip from pile From [1]: Amelia wins..

So, from [4]. Clinton tanding next to no scarred man and each of the other men tanding next to exactly one scarred man. Then either represents Clinton: IA no. C Ino yes no no yes yes Scarred? So, from [3], Case I is impossible. S ]n1e, and unscarred. In Case 11, the man next to Clinton cannot be handsome, [3]. Then, from [3], the man on the other end from Clinton be handsome. If Abraham is handsome, Barrett is on other end from [3]. If Abraham is not handsome, Barrett tanding next to Clinton from [3].

So either B represents and D represents Douglas: BAD C no no yes an Jsome? Then, from [I] and [2], the possible digits for N and A are as follows: So, choosing values for Rand L from [I], the table call be continued: So, from [2], Case vi I eliminated and, from [I], Cases i and v are eliminated.

Then continuing the table: So, from [1] and [2], ii is eliminated and S is So if one student got five answers, then each student got at least one correct wer. But then Betty and Doris would each exactly two correct answers, and Carol and Ellen would have exactly three correct answers, contradicting [2]. So one got five correct answers. Then the student who got none correct not be Adele, Carol, or Doris because the total number of answers cannot be the 'required 10 when the correct So Betty t Ellen got none correct.

If Betty got none correct, then the correct answers I the 'true-false questions would be: