T H _
M _ R C H _ N T
H _ S
3 2
G _ L D
C _ _ N S .
T _
V _ R _ F Y
T H _ S ,
D _ V _ D _
T H _
3 2
C _ _ N S
_ N T _
T W _
_ N _ Q _ _ L
N _ M B _ R S ,
S _ Y ,
2 7
_ N D
5 .
T H _ N :
3 2
( 2 7
-
5 )
=
( 2 7 2 )
-
( 5 2 ) .
7 0 4
=
7 2 9
-
2 5
I F
T H _
T W _
_ N _ Q _ _ L
N _ M B _ R S
_ R _
2 2
_ N D
1 0 ,
T H _ N :
3 2
( 2 2
-
1 0 )
=
( 2 2 2 )
-
( 1 0 2 ) .
3 8 4
=
4 8 4
-
1 0 0
A L T _ R N _ T _ V _ L Y ,
L _ T ' S
S _ Y
T H _ T
T H _
2
N _ M B _ R S
_ R _
X
_ N D
Y .
W _
C _ N
T H _ N
C _ M _
_ P
W _ T H
T H _
F _ L L _ W _ N G
_ Q _ _ T _ _ N :
3 2 ( X
-
Y )
=
X 2
-
Y 2
X 2
-
Y 2
C _ N
B _
_ X P _ N D _ D
T _ :
( X
-
Y ) ( X
+
Y )
S _ ,
N _ W
T H _
_ Q _ _ T _ _ N
W _ L L
B _ C _ M _ :
3 2
=
X
+
Y
( X
+
Y )
W _ L L
G _ V _
T H _
T _ T _ L
N _ M B _ R
_ F
G _ L D
C _ _ N S Clue
THE MERCHANT HAS 32 GOLD COINS. TO VERIFY THIS, DIVIDE THE 32 COINS INTO TWO UNEQUAL NUMBERS, SAY, 27 AND 5. THEN: 32 (27 - 5) = (272) - (52). 704 = 729 - 25 IF THE TWO UNEQUAL NUMBERS ARE 22 AND 10, THEN: 32 (22 - 10) = (222) - (102). 384 = 484 - 100 ALTERNATIVELY, LET'S SAY THAT THE 2 NUMBERS ARE X AND Y. WE CAN THEN COME UP WITH THE FOLLOWING EQUATION: 32(X - Y) = X2 - Y2 X2 - Y2 CAN BE EXPANDED TO: (X - Y)(X + Y) SO, NOW THE EQUATION WILL BECOME: 32 = X + Y (X + Y) WILL GIVE THE TOTAL NUMBER OF GOLD COINS THE CHILDREN ARE 1, 6 AND 6 YEARS OLD. THE PRODUCT OF THEIR AGES IS 36, SO NONE OF THEM CAN BE OLDER THAN 36. THE NUMBER 36 HAS TO BE EXPRESSED AS THE PRODUCT OF 3 NUMBERS. THEIR POSSIBLE AGES ARE (THE SUM OF THEIR AGES IS IN BRACKETS): 1, 1, 36 (3938) 1, 2, 18 (21) 1, 3, 12 (16) 1, 4, 9 (14) 1, 6, 6 (13) 2, 2, 9 (13) 2, 3, 6 (11) 3, 3, 4 (10) SINCE CHERYL IS TOM'S NEXT DOOR NEIGHBOUR, TOM KNOWS CHERYL'S HOUSE NUMBER. TOM WOULD KNOW THE CHILDREN'S AGES IN EVERY CASE THAT SUMS UP TO A UNIQUE NUMBER EXCEPT FOR THE SUM OF 13, WHICH HAVE 2 COMBINATIONS OF POSSIBLE AGES. AS A RESULT, TOM WOULD BE CONFUSED AS HE HAS TO PICK BETWEEN THE 2 COMBINATIONS: (1,6,6) AND (2,2,9). CHERYL THEN TELLS TOM ABOUT HER YOUNGEST CHILD WHO LIKES STRAWBERRY MILK WHICH TELLS TOM THAT THERE IS ONLY 1 YOUNGEST CHILD 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 THE IDOLS FROM LEFT TO RIGHT ARE: GOD OF DIPLOMACY, GOD OF FALSEHOOD, GOD OF TRUTH. THE GOD OF TRUTH IS NOT SEATED ON THE LEFT BECAUSE HE ALWAYS SPEAKS THE TRUTH WHEREAS THE IDOL ON THE LEFT REPLIED THAT THE GOD OF TRUTH IS SEATED AT THE CENTRE. THE GOD OF TRUTH IS ALSO NOT SEATED IN THE CENTRE AS HE ALWAYS SPEAKS THE TRUTH BUT THE IDOL AT THE CENTRE REPLIED THAT THE GOD OF DIPLOMACY IS SEATED AT THE CENTRE. THEREFORE, THE GOD OF TRUTH IS SEATED ON THE RIGHT. AS GOD OF TRUTH IS SEATED ON THE RIGHT, AND HE ALWAYS SPEAKS THE TRUTH, THEN THE THE GOD OF FALSEHOOD IS SEATED AT THE CENTRE. THE GOD OF DIPLOMACY IS SEATED ON THE LEFT AND HE HAS LIED