Brain Teaser No 3
Posted by: Don Atkinson on 11 March 2007
I know I started Brain Teaser No 1 about 5 years ago. ISTR another with No 2 in the title, so hopefully this is not duplicating somebody elses Brain Teaser No 3.....
Flight Around the World
A group of aeroplanes is based on a small island. Each plane holds just enough fuel to take it half way around the world. Any amount of fuel can be transfered from the tank of one aeroplane to another aeroplane whilst the planes are in flight. The ONLY source of fuel available to these aeroplanes is on this small island. Assume that there is no time lost when refueling, either in the air or on the ground.
What is the smallest number of aeroplanes required to ensure the flight of one aeroplane around the world on a great circle, assuming that all areoplanes have the same constant groundspeed and rate of fuel consumption and that all aeroplanes return safely to their island base.
Cheers
Don
Posted on: 28 September 2007 by Don Atkinson
Steve
You might want to take another look at 53 again?
I think it needs a slight tweak?
Cheers
Don
Posted on: 30 September 2007 by Don Atkinson
Steve,
Apologies
I somehow dropped the most important part of my message in the cut-and-paste stage which was "well done" even on the 53, however.....etc
56 = (4! - .4x4)/.4
Cheers
Don
Posted on: 01 October 2007 by steved
Don - sorry about the error in 53!
53 = 44 + 4/.4 (the point 4 has a dot above the 4).
an easier way to 56?
56 = 4! + 4! + 4 + 4
57 = 4! + 4! + 4/.4 (in the point 4, the 4 has a dot above it)
58 = 4! + 4! + 4/.4
Steve D
Posted on: 01 October 2007 by steved
59 = (4!/.4) - (4/4)
60 = (4!/.4) * (4/4)
61 = (4!/.4) + (4/4)
62 = (4!/.4) + 4 - sqrt(4)
Steve D
Posted on: 01 October 2007 by steved
63 = (4^4 - 4)/4 ie 4 to the 4th power, minus 4, all over 4
64 = 4! + 4! + (4*4)
65 = (4^4 + 4)/4
66 = 4^4/4 + sqrt(4)
Steve D
Posted on: 01 October 2007 by steved
67 = (4! + sqrt(4))/.4 + sqrt(4)
68 = 4!/.4 + 4 + 4
69 = four-play!!
Steve D
Posted on: 02 October 2007 by Don Atkinson
quote:
69 = four-play!!
Cheers
Don
Posted on: 06 October 2007 by Matthew T
A more mathetical approach to 69
69 = 4!/.4 + 4/.4(dot on top)
70 = 4!! x 4!! + 4 + sqrt(4)
Posted on: 16 October 2007 by Matthew T
71 = 4! x (4!/4!!) - int(log(4!))
72 = 4! + 44 + 4
73 = 4! x (4!/4!!) + int(log(4!))
74 = 4! x (4!/4!!) + sqrt(4)
Posted on: 16 October 2007 by Don Atkinson
!!!!!!!!!!!!!!
Cheers
Don
Posted on: 16 October 2007 by Don Atkinson
A spate of .4[with a dot above]
75 = ((4!sqrt4) + sqrt4)/sqrt.4[with a dot above]
76 = 4 + (4!x sqrt(4/.4[with a dot above]))
77 = ((4/.4[with a dot above])^sqrt4) - 4
78 = (4! + 4!(.4[with a dot above]))/.4[with a dot above]
79 = ((4/.4[with a dot above])^sqrt4) - sqrt4
Cheers
Don
Posted on: 17 October 2007 by Alexander
80= 4! * 4 - 4 * 4
Posted on: 18 October 2007 by Matthew T
81 = 4 x 4 / (.4 with dot x .4 with dot)
82 = 4!! x 4/.4 + sqrt4
83 = (4!!)! / (4!!)!! - 4! + sqrt4
84 = $!! x 4/.4 +4
85 = (4!!)! / (4!!)!! - 4! + 4
Posted on: 18 October 2007 by Don Atkinson
quote:
84 = $!! x 4/.4 +4
Try as you might, there is still NO PRIZE MONEY on this thread
Cheers
Don
Posted on: 28 October 2007 by Don Atkinson
86 = 4!4 - (4/.4)
87 = 4!4 - (4/.4[with a dot above])
88 = 4!4 - 4sqrt4
89 = ((4! + sqrt4)/.4) + 4!
90 = 4.4/(.4)(.4[with a dot above]) (the first part is four point four, not four times four)
Cheers
Don
Posted on: 28 October 2007 by Don Atkinson
.
Posted on: 28 October 2007 by Alexander
91 = 4! * (4-inv(4) + inv(4!))
with inv(n)= 1/n
I hate your dots, Don
Posted on: 28 October 2007 by Alexander
92 = 4!*4 -sqrt(4) -sqrt(4)
93 = 4!*4 - 4!/4!!
94 = 4!*4 - 4!!/4
95 = 4!*4 - 4/4
96 = 4!*4 - 4 + 4
and it's a straight line to the finish...
Posted on: 28 October 2007 by Don Atkinson
quote:
I hate your dots, Don
I posted one above, especially for you!!!
Cheers
Don
Posted on: 28 October 2007 by Don Atkinson
97 = 4!*4 + 4/4
98 = 4!*4 + 4/sqrt4
99 = 4!*4 + sqrt((4/.4[with a dot above])
100 = ......would someone else care to do the honours?
Cheers
Don
Posted on: 28 October 2007 by Alexander
quote:
Originally posted by Don Atkinson:
quote:
I hate your dots, Don
I posted one above, especially for you!!!
Cheers
Don
Thanks Don, that dot was a nice dot, with good compact dottiness.
Alexander
Posted on: 29 October 2007 by Matthew T
100 = 4! x 4 + 4!! - 4
Posted on: 02 November 2007 by Matthew T
Three men, members of a safari, are captured by cannibals in the jungle. The men are given one chance to escape with their lives. The men are lined up and bound to stakes such that one man can see the backs of the other two, the middle man can see the back of the front man, and the front man can't see anybody. The men are shown five hats, three of which are black and two of which are white. Then the men are blindfolded, and one of the five hats is placed on each man's head. The remaining two hats are hidden away. The blindfolds are removed. The men are told that if just one of the men can guess what hat he's wearing, they may all go free. Time passes. Finally, the front man, who can't see anyone, correctly guesses the color of his hat. What color was it, and how did he guess correctly?
Posted on: 02 November 2007 by Alexander
Black. but let's see if anyone else finds the reasoning. I'm assuming that the three men aren't completely daft. I'm also assuming the three men don't know that the cannibals aren't being completely honnist and they're still going to make a delicious stew with them, with carrots and turnips and celeriac and a modicum of panache.
Posted on: 02 November 2007 by Ian G.
Black ? - and he didn't so much guess as depend on the rational behaviour of his comrades - I think.
