If Cab stands guard on Wednesday, then Art tands on Thursday. Then Bob stands guard on Tuesday and. Then Cab stands on Saturday. 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 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.
Ie:[Onls--contradicting previous reasoning. 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. So Case 1 is the t one and, from [I], Cass is the widow. 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: P or D. 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: 4, 8, 12, 16, 20, 24, From [3], Clyde has at least 5 children and a number 01 children from this sequence : 5, 7, 9, II , 13, 15, 17, 19, 21 , 23, Then the total number of children is at least 12 and, from [41 at most Also : 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. 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: t09 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.
Ay,:- won one game, Mrs. 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. So eithcl Adam or Dawn occupies dressing room Il. Then, from [3], one of the following sets of occupation must exist: from [2], the first of these sets is impossible. 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. 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 : C A? 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, from [2], Case vi I eliminated and, from [I], Cases i and v are eliminated.
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 : Ill. We like to set the bar high, roll up our sleeves and work together to out-perform the competition.
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