T H _
S H _ _ P
W _ L L
S _ R V _ V _ .
I F
T H _ R _
W _ R _
1
L _ _ N
_ N D
1
S H _ _ P ,
T H _ N
T H _
L _ _ N
W _ _ L D
S _ M P L Y
_ _ T
T H _
S H _ _ P .
T H _
S H _ _ P
W _ L L
N _ T
S _ R V _ V _ .
I F
T H _ R _
W _ R _
2
L _ _ N S
_ N D
1
S H _ _ P ,
T H _ N
N _
L _ _ N
W _ _ L D
_ _ T
T H _
S H _ _ P ,
B _ C _ _ S _
_ F
_ N _
_ F
T H _ M
W _ _ L D ,
_ T
W _ _ L D
S _ R _ L Y
B _
_ _ T _ N
B Y
T H _
_ T H _ R
L _ _ N
_ F T _ R W _ R D S .
T H _
S H _ _ P
W _ L L
S _ R V _ V _ .
I F
T H _ R _
W _ R _
3
L _ _ N S
_ N D
1
S H _ _ P ,
T H _ N
_ N _
_ F
T H _
L _ _ N S
C _ _ L D
S _ F _ L Y
_ _ T
T H _
S H _ _ P ,
B _ C _ _ S _
_ T
W _ _ L D
T _ R N
_ N T _
T H _
S C _ N _ R _ _
W _ T H
2
L _ _ N S ,
W H _ R _
N _
_ N _
C _ N
_ _ T
T H _
S H _ _ P .
T H _
S H _ _ P
W _ L L
N _ T
S _ R V _ V _ .
I F
T H _ R _
W _ R _
4
L _ _ N S
_ N D
1
S H _ _ P ,
T H _ N
N _
L _ _ N
W _ _ L D
_ _ T
T H _
S H _ _ P ,
B _ C _ _ S _
_ T
W _ _ L D
T _ R N
_ N T _
T H _
S C _ N _ R _ _
W _ T H
3
L _ _ N S .
T H _
S H _ _ P
W _ L L
S _ R V _ V _ .
C _ N T _ N _ _ N G
T H _ S
_ R G _ M _ N T ,
T H _
C _ N C L _ S _ _ N
_ S
_ S
F _ L L _ W S :
I F
T H _ R _
_ S
_ N
_ V _ N
N _ M B _ R
_ F
L _ _ N S ,
T H _ N
N _ T H _ N G
H _ P P _ N S
_ N D
T H _
S H _ _ P
S _ R V _ V _ S .
I F
T H _ R _
_ S
_ N
_ D D
N _ M B _ R
_ F
L _ _ N S ,
T H _ N
_ N Y
L _ _ N
C _ _ L D
S _ F _ L Y
_ _ T
T H _
S H _ _ P
_ N D
T H _
S H _ _ P
W _ L L
N _ T
S _ R V _ V _ .
T H _ S
_ S
S _ M _ L _ R
T _
T H _
U N _ X P _ C T _ D
H _ N G _ N G
P _ R _ D _ X Clue
THE PROBLEM IS THAT THE QUESTION IS CLEVERLY PHRASED TO CAUSE CONFUSION. THE BALANCE TOTAL DOES NOT NEED TO BE EQUAL TO THE WITHDRAWAL TOTAL. AND IN THIS CASE, THE BALANCE TOTAL WILL NOT ADD UP TO SOMETHING MEANINGFUL. LET'S SAY YOU HAVE $100 IN YOUR BANK ACCOUNT AND YOU MAKE 2 WITHDRAWALS OF $1 EACH. THE WITHDRAWAL TOTAL IS $2 ($1 + $1). BUT THE BALANCE TOTAL IS $197 ($99 + $98) THE MAXIMUM NUMBER OF BANANAS THAT CAN BE TRANSFERRED IS 533. IF WE TRANSPORT 1000 BANANAS AT A TIME, THE CAMEL WILL CONSUME ALL THE BANANAS BY THE TIME IT REACHES THE DESTINATION. SO, WE NEED TO HAVE INTERMEDIATE DROP POINTS, THE CAMEL CAN THEN MAKE SEVERAL SHORT TRIPS IN BETWEEN. TO BE OPTIMAL, WE TRY TO MAINTAIN THE NUMBER OF BANANAS AT EACH POINT TO BE A MULTIPLE OF 1000, AS THAT'S THE MAXIMUM OF BANANAS THE CAMEL CAN TRANSPORT AT ANY POINT OF TIME. SOURCE---IP1---IP2----DESTINATION 3000 X KM 2000 Y KM 1000 Z KM TO GO FROM SOURCE TO IP1 POINT CAMEL HAS TO TAKE A TOTAL OF 5 TRIPS, 3 FORWARD AND 2 BACKWARDS, SINCE WE HAVE 3000 BANANAS TO TRANSPORT. THE SAME WAY FROM IP1 TO IP2 CAMEL HAS TO TAKE A TOTAL OF 3 TRIPS, 2 FORWARD AND 1 BACKWARD, SINCE WE HAVE 2000 BANANAS TO TRANSPORT. FROM IP2 TO DESTINATION WE ONLY HAVE 1 FORWARD MOVE. LET'S SEE THE TOTAL NUMBER OF BANANAS CONSUMED AT EVERY POINT. FROM THE SOURCE TO IP1 ITS 5X BANANAS, AS THE DISTANCE BETWEEN THE SOURCE AND IP1 IS X KM AND THE CAMEL HAD 5 TRIPS. FROM IP1 TO IP2 ITS 3Y BANANAS, AS THE DISTANCE BETWEEN IP1 AND IP2 IS Y KM AND THE CAMEL HAD 3 TRIPS. FROM IP2 TO DESTINATION ITS Z BANANAS. WE CAN NOW CALCULATE THE DISTANCE BETWEEN THE POINTS: 3000 - 5X = 2000 SO WE GET X = 200 2000-3Y = 1000 SO WE GET Y = 333.33 BUT HERE THE DISTANCE IS ALSO THE NUMBER OF BANANAS AND IT CANNOT BE FRACTION SO WE TAKE Y = 333 AND AT IP2 WE HAVE THE NUMBER OF BANANAS EQUAL 1001, SO ITS 2000-3Y = 1001 SO THE REMAINING DISTANCE TO THE MARKET IS 1000 - X - Y = Z I.E 1000-200-333 = Z = 467. FROM IP2 TO THE DESTINATION POINT, THE CAMEL CONSUMES 467 BANANAS AND 533 BANANAS REMAIN. REFERENCE: A CAMEL TRANSPORTING BANANAS - PUZZLING STACK EXCHANGE IF THERE WAS ONLY ONE BLUE-EYED PERSON ON THE ISLAND, THEN THAT PERSON WOULD LOOK AROUND AND SEE THAT THERE IS NO OTHER BLUE-EYED PERSON. SO HE REALIZES THAT HE IS THE ONLY PERSON WITH BLUE EYES ON THE ISLAND AND LEAVES ON THE DAY OF THE ANNOUNCEMENT. IF THERE ARE 2 BLUE-EYED PEOPLE, THEN THEY LOOK AT EACH OTHER. EACH ONE EXPECTS THE OTHER TO LEAVE ON THE DAY OF THE ANNOUNCEMENT. HOWEVER, ON THE NEXT DAY, WHEN THEY REALIZE THAT NEITHER OF THEM LEFT THE ISLAND, THEY WOULD BE ABLE TO DEDUCE THAT BOTH OF THEM HAVE BLUE EYES. THEY BOTH LEAVE THE ISLAND ON THE SECOND DAY. THROUGH MATHEMATICAL INDUCTION, THIS LOGIC CAN BE APPLIED TO THE 100 BLUE-EYED PEOPLE ON THE ISLAND. SO ON THE 100TH DAY, ALL THE 100 BLUE-EYED PEOPLE LEAVE THE ISLAND THE SHEEP WILL SURVIVE. IF THERE WERE 1 LION AND 1 SHEEP, THEN THE LION WOULD SIMPLY EAT THE SHEEP. THE SHEEP WILL NOT SURVIVE. IF THERE WERE 2 LIONS AND 1 SHEEP, THEN NO LION WOULD EAT THE SHEEP, BECAUSE IF ONE OF THEM WOULD, IT WOULD SURELY BE EATEN BY THE OTHER LION AFTERWARDS. THE SHEEP WILL SURVIVE. IF THERE WERE 3 LIONS AND 1 SHEEP, THEN ONE OF THE LIONS COULD SAFELY EAT THE SHEEP, BECAUSE IT WOULD TURN INTO THE SCENARIO WITH 2 LIONS, WHERE NO ONE CAN EAT THE SHEEP. THE SHEEP WILL NOT SURVIVE. IF THERE WERE 4 LIONS AND 1 SHEEP, THEN NO LION WOULD EAT THE SHEEP, BECAUSE IT WOULD TURN INTO THE SCENARIO WITH 3 LIONS. THE SHEEP WILL SURVIVE. CONTINUING THIS ARGUMENT, THE CONCLUSION IS AS FOLLOWS: IF THERE IS AN EVEN NUMBER OF LIONS, THEN NOTHING HAPPENS AND THE SHEEP SURVIVES. IF THERE IS AN ODD NUMBER OF LIONS, THEN ANY LION COULD SAFELY EAT THE SHEEP AND THE SHEEP WILL NOT SURVIVE. THIS IS SIMILAR TO THE UNEXPECTED HANGING PARADOX