2 5
_ G G S
W _ R _
B R _ K _ N .
T H _ R _
_ S
_ N L Y
_ N _
W _ Y
_ F
F _ N D _ N G
_
S _ L _ T _ _ N
T _
T H _ S
P R _ B L _ M :
N _ M B _ R S
W H _ C H
L _ _ V _
_
R _ M _ _ N D _ R
_ F
1 ,
W H _ N
D _ V _ D _ D
B Y
2 :
3 ,
5 ,
7 ,
9 ,
1 1 ,
1 3 ,
1 5 ,
1 7 ,
1 9 ,
2 1 ,
2 3 ,
2 5 ,
2 7 ,
2 9 ,
3 1 ,
3 3 ,
3 5
N _ M B _ R S
W H _ C H
L _ _ V _
_
R _ M _ _ N D _ R
_ F
1 ,
W H _ N
D _ V _ D _ D
B Y
3 :
4 ,
7 ,
1 0 ,
1 3 ,
1 6 ,
1 9 ,
2 2 ,
2 5 ,
2 8 ,
3 1 ,
3 4 ,
3 7
N _ M B _ R S
W H _ C H
L _ _ V _
_
R _ M _ _ N D _ R
_ F
1 ,
W H _ N
D _ V _ D _ D
B Y
4 :
5 ,
9 ,
1 3 ,
1 7 ,
2 1 ,
2 5 ,
2 9 ,
3 3 ,
3 7
N _ M B _ R S
W H _ C H
L _ _ V _
N _
R _ M _ _ N D _ R
W H _ N
D _ V _ D _ D
B Y
5 :
5 ,
1 0 ,
1 5 ,
2 0 ,
2 5 ,
3 0 ,
3 5
T H _
_ N L Y
N _ M B _ R
F _ L F _ L L _ N G
T H _
F _ _ R
C _ N D _ T _ _ N S
_ S
2 5 Clue
25 EGGS WERE BROKEN. THERE IS ONLY ONE WAY OF FINDING A SOLUTION TO THIS PROBLEM: NUMBERS WHICH LEAVE A REMAINDER OF 1, WHEN DIVIDED BY 2: 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35 NUMBERS WHICH LEAVE A REMAINDER OF 1, WHEN DIVIDED BY 3: 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37 NUMBERS WHICH LEAVE A REMAINDER OF 1, WHEN DIVIDED BY 4: 5, 9, 13, 17, 21, 25, 29, 33, 37 NUMBERS WHICH LEAVE NO REMAINDER WHEN DIVIDED BY 5: 5, 10, 15, 20, 25, 30, 35 THE ONLY NUMBER FULFILLING THE FOUR CONDITIONS IS 25 HE HAD 15 EGGS. THE FIRST PERSON BOUGHT HALF OF THE 15 EGGS WHICH IS 7½ EGGS AND ANOTHER ½ EGG. SO HE BOUGHT 8 EGGS IN TOTAL AND JACK IS LEFT WITH 7 EGGS. THE SECOND PERSON BOUGHT HALF OF THE REMAINING 7 EGGS WHICH IS 3½ EGGS AND ANOTHER ½ EGG. SO HE BOUGHT 4 EGGS AND JACK IS LEFT WITH 3 EGGS. THE LAST PERSON BOUGHT HALF OF THE REMAINING 3 EGGS WHICH IS 1½ EGGS AND ANOTHER ½ EGG. SO THE LAST PERSON BOUGHT 2 EGGS. THAT LEAVES 1 EGG REMAINING IN JACK'S BASKET THE MESSENGER HAS TO HAVE TRAVELED 2 KM. IT DOESN'T MATTER WHAT SPEED THEY WALKED AT. AT THE BEGINNING OF THE PUZZLE, THE LINE IS 1 KM LONG. THE GENERAL IS THEREFORE 1 KM AHEAD OF HIM. THE MESSENGER MUST THEREFORE TRAVEL MORE THAN 1 KM TO REACH THE GENERAL. SINCE THE LINE MOVES 1 KM FORWARD, THE END IS WHERE THE BEGINNING WAS. EVEN IF HE WALKED 1.5 KM TO THE GENERAL, HE ONLY HAS TO WALK 0.5 KM TO GET BACK TO THE END OF THE LINE. IT GOES FASTER GOING BACK, BECAUSE NOW THEY ARE COMING TOWARDS HIM, AND NOT GOING AWAY 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