Z _ _ ' S
S M _ L L _ S T
P _ S S _ B L _
N _ M B _ R
_ S
6 .
B _ S _ D
_ N
T H _
F _ R S T
S T _ T _ M _ N T
_ F
A L _ ,
_ T
_ N D _ C _ T _ S
T H _ T
H _
H _ S
N _ _ T H _ R
1
N _ R
9 .
I F
H _
H _ D
_ _ T H _ R
1
_ R
9
T H _ N
H _
W _ _ L D
K N _ W
T H _ T
Z _ _
M _ S T
H _ V _
_
B _ G G _ R
_ R
S M _ L L _ R
N _ M B _ R .
N _ W
Z _ _ ,
B _ S _ D
_ N
A L _ ' S
F _ R S T
S T _ T _ M _ N T ,
K N _ W S
T H _ T
A L _
D _ _ S N ' T
H _ V _
1
_ R
9 .
Z _ _ ' S
F _ R S T
S T _ T _ M _ N T
_ N D _ C _ T _ S
T H _ T
S H _
D _ _ S
N _ T
H _ V _
2
_ R
8
( N _ _ T H _ R
1
N _ R
9 ) .
I F
S H _
H _ D
1 ,
2 ,
8
_ R
9 ,
T H _ N
S H _
C _ _ L D
H _ V _
C _ N C L _ D _ D
T H _ T
A L _
H _ S
_
B _ G G _ R
_ R
S M _ L L _ R
N _ M B _ R .
N _ W
A L _
K N _ W S
T H _ T
Z _ _
D _ _ S N ' T
H _ V _
1 ,
2 ,
8
_ R
9 .
A L _ ' S
S _ C _ N D
S T _ T _ M _ N T
_ N D _ C _ T _ S
T H _ T
H _
D _ _ S
N _ T
H _ V _
3
_ R
7
_ N D
_ L S _
N _ T
1 ,
2 ,
8
_ R
9 .
Z _ N _
C _ N
C _ N C L _ D _
T H _ T
A L _
D _ _ S N ' T
H _ V _
1 ,
2 ,
3 ,
7 ,
8
_ R
9 .
I N
S H _ R T ,
A L _
M _ S T
H _ V _
_ _ T H _ R
4 ,
5
_ R
6 .
N _ W
W H _ N
Z _ _
S _ Y S
T H _ T
S H _
H _ S
_
B _ G G _ R
N _ M B _ R
T H _ N
_ T
M _ S T
B _
_ _ T H _ R
6 ,
7 ,
8
_ R
9
_ N D
A L _
H _ V _ N G
4
_ R
5 .
Z _ _
C _ N ' T
S _ Y
C _ N F _ D _ N T L Y
T H _ T
S H _
H _ S
_
B _ G G _ R
N _ M B _ R
_ F
S H _
H _ D
_
4
_ R
5 ,
_ S
_ T
C _ _ L D
B _
S M _ L L _ R
T H _ N
W H _ T
A L _
C _ _ L D
H _ V _ .
S _
Z _ _ ' S
S M _ L L _ S T
P _ S S _ B L _
N _ M B _ R
_ S
_
6 Clue
FIRST WEIGHING: FOUR AGAINST FOUR SECOND WEIGHING: TWO AGAINST TWO THIRD WEIGHING: ONE AGAINST ONE LET'S NAME THE BALLS 1-12. FIRST WE WEIGH {1,2,3,4} ON THE LEFT AND {5,6,7,8} ON THE RIGHT. THERE ARE THREE SCENARIOS WHICH CAN ARISE FROM THIS. IF THEY BALANCE, THEN WE KNOW 9, 10, 11 OR 12 IS ODD. WEIGH {8, 9} AND {10, 11} (NOTE: 8 IS NOT ODD) IF THEY BALANCE, WE KNOW 12 IS THE ODD ONE. JUST WEIGH IT WITH ANY OTHER BALL AND FIGURE OUT IF IT IS LIGHTER OR HEAVIER. IF {8, 9} IS HEAVIER, THEN EITHER 9 IS HEAVY OR 10 IS LIGHT OR 11 IS LIGHT. WEIGH {10} AND {11}. IF THEY BALANCE, 9 IS ODD (HEAVIER). IF THEY DON'T BALANCE THEN WHICHEVER ONE IS LIGHTER IS ODD (LIGHTER). IF {8, 9} IS LIGHTER, THEN EITHER 9 IS LIGHT OR 10 IS HEAVY OR 11 IS HEAVY. WEIGH {10} AND {11}. IF THEY BALANCE, 9 IS ODD (LIGHTER). IF THEY DON'T BALANCE THEN WHICHEVER ONE IS HEAVIER IS ODD (HEAVIER). IF {1,2,3,4} IS HEAVIER, WE KNOW EITHER ONE OF {1,2,3,4} HEAVIER OR ONE OF {5,6,7,8} IS LIGHTER BUT IT IS GUARANTEED THAT {9,10,11,12} ARE NOT ODD. WEIGH {1,2,5} AND {3,6,9} (NOTE: 9 IS NOT ODD). IF THEY BALANCE, THEN EITHER 4 IS HEAVY OR 7 IS LIGHT OR 8 IS LIGHT. FOLLOWING THE LAST STEP FROM THE PREVIOUS CASE, WE WEIGH {7} AND {8}. IF THEY BALANCE, 4 IS ODD (HEAVIER). IF THEY DON'T BALANCE THEN WHICHEVER ONE IS LIGHTER IS ODD (LIGHTER). IF {1,2,5} IS HEAVIER, THEN EITHER 1 IS HEAVY OR 2 IS HEAVY OR 6 IS LIGHT. WEIGH {1} AND {2}. IF THEY BALANCE, 6 IS ODD (LIGHTER). IF THEY DON'T BALANCE THEN WHICHEVER ONE IS HEAVIER IS ODD (HEAVIER). IF {3,6,9} IS HEAVIER, THEN EITHER 3 IS HEAVY OR 5 IS LIGHT. WEIGH {5} AND {9}. THEY WON'T BALANCE. IF {5} IS LIGHTER, 5 IS ODD (LIGHTER). IF THEY BALANCE, 3 IS ODD (HEAVIER). IF {5,6,7,8} IS HEAVIER, IT IS THE SAME SITUATION AS IF {1,2,3,4} WAS HEAVIER. JUST PERFORM THE SAME STEPS USING 5,6,7 AND 8. WEIGH {5,6,1} AND {7,2,9} (NOTE: 9 IS NOT ODD). IF THEY BALANCE, THEN EITHER 8 IS HEAVY OR 3 IS LIGHT OR 4 IS LIGHT. WE WEIGH {3} AND {4}. IF THEY BALANCE, 8 IS ODD (HEAVIER). IF THEY DON'T BALANCE THEN WHICHEVER ONE IS LIGHTER IS ODD (LIGHTER). IF {5,6,1} IS HEAVIER, THEN EITHER 5 IS HEAVY OR 6 IS HEAVY OR 2 IS LIGHT. WEIGH {5} AND {6}. IF THEY BALANCE, 2 IS ODD (LIGHTER). IF THEY DON'T BALANCE THEN WHICHEVER ONE IS HEAVIER IS ODD (HEAVIER). IF {7,2,9} IS HEAVIER, THEN EITHER 7 IS HEAVY OR 1 IS LIGHT. WEIGH {1} AND {9}. IF THEY BALANCE, 7 IS ODD (HEAVIER). IF THEY DON'T BALANCE THEN 1 IS ODD (LIGHTER). NOTE: THERE ARE OTHER POSSIBLE SOLUTIONS TO THIS PROBLEM AS WELL FIRST HEAP: 22, SECOND HEAP: 14, THIRD HEAP: 12. TO SOLVE THIS PROBLEM, WE SHALL HAVE TO START FROM THE END. WE HAVE BEEN TOLD THAT AFTER ALL THE TRANSPOSITIONS, THE NUMBER OF MATCHES IN EACH HEAP IS THE SAME. LET US PROCEED FROM THIS FACT. SINCE THE TOTAL NUMBER OF MATCHES HAS NOT CHANGED IN THE PROCESS, AND THE TOTAL NUMBER BEING 48, IT FOLLOWS THAT THERE WERE 16 MATCHES IN EACH HEAP. AND SO, IN THE END WE HAVE: FIRST HEAP: 16, SECOND HEAP: 16, THIRD HEAP: 16 IMMEDIATELY BEFORE THIS WE HAVE ADDED TO THE FIRST HEAP AS MANY MATCHES AS THERE WERE IN IT, I.E. WE HAD DOUBLED THE NUMBER. SO, BEFORE THE FINAL TRANSPOSITION, THERE ARE ONLY 8 MATCHES IN THE FIRST HEAP. NOW, IN THE THIRD HEAP, FROM WHICH WE TOOK THESE 8 MATCHES, THERE WERE: 16 + 8 = 24 MATCHES. WE NOW HAVE THE NUMBERS AS FOLLOWS: FIRST HEAP: 8, SECOND HEAP: 16, THIRD HEAP: 24. WE KNOW THAT WE TOOK FROM THE SECOND HEAP AS MANY MATCHES AS THERE WERE IN THE THIRD HEAP, WHICH MEANS 24 WAS DOUBLE THE ORIGINAL NUMBER. FROM THIS WE KNOW HOW MANY MATCHES WE HAD IN EACH HEAP AFTER THE FIRST TRANSPOSITION: FIRST HEAP: 8, SECOND HEAP: 16 + 12 = 28, THIRD HEAP: 12. NOW WE CAN DRAW THE FINAL CONCLUSION THAT BEFORE THE FIRST TRANSPOSITION THE NUMBER OF MATCHES IN EACH HEAP WAS: FIRST HEAP: 22, SECOND HEAP: 14, THIRD HEAP: 12 7 RACES. LET'S NAME THE RACES R1 THROUGH R7. LET RXN REPRESENT A HORSE IN RACE X, FINISHING IN NTH PLACE. SO R32 REPRESENTS A HORSE THAT FINISHED 2ND PLACE IN THE 3RD RACE. GROUP THE 25 HORSES INTO 5 GROUPS OF 5 AND RACE THEM. THE 4TH AND 5TH PLACED HORSES OF EACH RACE CAN BE ELIMINATED SINCE THEY CANNOT MEET THE CRITERIA OF 3 FASTEST HORSES. WE ARE NOW LEFT WITH 15 HORSES (5 GROUPS OF 3 HORSES); 3 HORSES FROM EACH RACE. FOR THE 6TH RACE, RACE THE FASTEST HORSE (R11, R21, R31, R41, R51) FROM EACH OF THE FIRST 5 RACES. THE WINNER OF THE 6TH RACE IS THE FASTEST HORSE. THE 4TH AND 5TH PLACED HORSES FROM THE 6TH RACE CAN BE ELIMINATED INCLUDING ALL THE HORSES WITHIN THEIR RESPECTIVE GROUPS. FOR EXAMPLE, IF THE HORSE THAT CAME IN 4TH PLACE IS FROM R41, THE HORSES R42 AND R43 CAN BE ELIMINATED AS WELL. LET'S SAY FOR THE 6TH RACE, R11 CAME FIRST, R21 CAME 2ND AND R31 CAME 3RD. WE KNOW THAT R11 IS THE FASTEST HORSE. WE NOW NEED TO DETERMINE THE 2ND AND 3RD FASTEST HORSES. WE CAN NOW ALSO ELIMINATE THE HORSES R23, R32 AND R33. THIS WILL LEAVE US WITH FIVE HORSES FOR THE 7TH RACE - R12, R13, R21, R22 AND R31. THE 1ST AND 2ND PLACED HORSES IN THE 7TH RACE ARE THE 2ND AND 3RD FASTEST HORSES ZOE'S SMALLEST POSSIBLE NUMBER IS 6. BASED ON THE FIRST STATEMENT OF ALI, IT INDICATES THAT HE HAS NEITHER 1 NOR 9. IF HE HAD EITHER 1 OR 9 THEN HE WOULD KNOW THAT ZOE MUST HAVE A BIGGER OR SMALLER NUMBER. NOW ZOE, BASED ON ALI'S FIRST STATEMENT, KNOWS THAT ALI DOESN'T HAVE 1 OR 9. ZOE'S FIRST STATEMENT INDICATES THAT SHE DOES NOT HAVE 2 OR 8 (NEITHER 1 NOR 9). IF SHE HAD 1, 2, 8 OR 9, THEN SHE COULD HAVE CONCLUDED THAT ALI HAS A BIGGER OR SMALLER NUMBER. NOW ALI KNOWS THAT ZOE DOESN'T HAVE 1, 2, 8 OR 9. ALI'S SECOND STATEMENT INDICATES THAT HE DOES NOT HAVE 3 OR 7 AND ALSO NOT 1, 2, 8 OR 9. ZONE CAN CONCLUDE THAT ALI DOESN'T HAVE 1, 2, 3, 7, 8 OR 9. IN SHORT, ALI MUST HAVE EITHER 4, 5 OR 6. NOW WHEN ZOE SAYS THAT SHE HAS A BIGGER NUMBER THEN IT MUST BE EITHER 6, 7, 8 OR 9 AND ALI HAVING 4 OR 5. ZOE CAN'T SAY CONFIDENTLY THAT SHE HAS A BIGGER NUMBER IF SHE HAD A 4 OR 5, AS IT COULD BE SMALLER THAN WHAT ALI COULD HAVE. SO ZOE'S SMALLEST POSSIBLE NUMBER IS A 6