Brain Teaser No 1
Posted by: Don Atkinson on 16 November 2001
THE EXPLORER
An explorer set off on a journey. He walked a mile south, a mile east and a mile north. At this point he was back at his start. Where on earth was his starting point? OK, other than the North Pole, which is pretty obvious, where else could he have started this journey?
Cheers
Don
Posted on: 22 April 2003 by Dan M
Don,
Your figure is almost an exact copy of mine (yours is a lot neater). Thanks for the help,
cheers
Dan
Your figure is almost an exact copy of mine (yours is a lot neater). Thanks for the help,
cheers
Dan
Posted on: 24 April 2003 by Matthew T
quote:
Originally posted by Don Atkinson:
A Monumental Problem.
A 10 tonne statue with a flat bottom, has to be placed upon a pedestal by crane. There is no way of lifting the statue except by slings under the base.
How could you arrange to set the statue on the pedestal and get the slings out ?
Cheers
Don
Still no solutions to this one, nothing springs to mind, anybody else got a clue?
Matthew
Posted on: 24 April 2003 by Dobbin
I posted
ice ice baby
some time ago
ice ice baby
some time ago
Posted on: 24 April 2003 by Don Atkinson
Matthew,
Dobbin said I posted ice ice baby some time ago
and I commented Cool answer man, cool answer.
Sorry it was all too subtle........
Just to remove the subtleness, a few suitable ice cubes, strategically placed on the plinth, statute lowered onto ice cube, slings withdrawn, ice cubes melt, little bit of panic as statute tilts slightly. and we're home and dry.......well, once the sun has dried the melted ice...
Cheers
Don
Dobbin said I posted ice ice baby some time ago
and I commented Cool answer man, cool answer.
Sorry it was all too subtle........
Just to remove the subtleness, a few suitable ice cubes, strategically placed on the plinth, statute lowered onto ice cube, slings withdrawn, ice cubes melt, little bit of panic as statute tilts slightly. and we're home and dry.......well, once the sun has dried the melted ice...
Cheers
Don
Posted on: 24 April 2003 by Don Atkinson
Dan,
p.s. so Don is 1/2 way the correct answer?
Sorry, didn't think to close this one out because both you and Matthew hit it hard.
But just to avoid doubt, 1/2 way was my view also..........
Unless someone else knows better....
Cheers
Don
p.s. so Don is 1/2 way the correct answer?
Sorry, didn't think to close this one out because both you and Matthew hit it hard.
But just to avoid doubt, 1/2 way was my view also..........
Unless someone else knows better....
Cheers
Don
Posted on: 24 April 2003 by Dan M
another simple (in the head) one:
After peeling an orange, I notice all the segments are identical. I eat two segments and realise that I haven't change the area of the peeled orange at all. How many segments are left?
-Dan
After peeling an orange, I notice all the segments are identical. I eat two segments and realise that I haven't change the area of the peeled orange at all. How many segments are left?
-Dan
Posted on: 25 April 2003 by Matthew T
six
The ice slipped me by!
Matthew
The ice slipped me by!
Matthew
Posted on: 25 April 2003 by Matthew T
A clock consists of 12 small identical circular plates, numbered and seated on a circle so that each touches both its neighbours. The clock also has a wheel that runs around the outside of the clockface, has the same radius as each of the small plates, and has its centre attached by a rod and spring to the minute hand. The spring keeps tension on the wheel in the direction of the clock centre so that as the minute hand rotates the wheel rolls along the outer contours of the 12 plates without losing surface contact. How many times does the wheel rotate with respect to the clock centre each hour?
Posted on: 25 April 2003 by Dan M
Matthew,
Six is correct.
I believe your wheel rotates 3 times.
-Dan
Six is correct.
I believe your wheel rotates 3 times.
-Dan
Posted on: 27 April 2003 by Don Atkinson
Ooooops,
Prompted by your post above, Lynn, i've just looked back at my 3/4/5 triangle drawing. Several times i have refered to 'similar' triangles when i really meant 'congruent' triangles.
100 times
I really meant congruent triangles
I really meant congruent triangles
I really meant congruent triangles
I really meant congruent tria.....
Cheers
Don
Prompted by your post above, Lynn, i've just looked back at my 3/4/5 triangle drawing. Several times i have refered to 'similar' triangles when i really meant 'congruent' triangles.
100 times
I really meant congruent triangles
I really meant congruent triangles
I really meant congruent triangles
I really meant congruent tria.....
Cheers
Don
Posted on: 28 April 2003 by Matthew T
Dan,
I think you will need to revisit the problem.
Matthew
I think you will need to revisit the problem.
Matthew
Posted on: 28 April 2003 by steved
Matthew,
I reckon the wheel rotates 4 times.
Steve D
I reckon the wheel rotates 4 times.
Steve D
Posted on: 28 April 2003 by Matthew T
Nope
Posted on: 28 April 2003 by JeremyD
quote:
How many times does the wheel rotate with respect to the clock centre each hour?
Once.
--J
Posted on: 28 April 2003 by Don Atkinson
So we're all agreed then?
The wheel rotates three times per hour; but with respect to the centre of the clock this is only once an hour.....
Well spotted Jeremy.
Dan, IMHO you're entitled to feel cheated. But as they say in all exams, you must read the question carefully. I'm sure Matthew will apologise....
Steve??????
Cheers
Don
The wheel rotates three times per hour; but with respect to the centre of the clock this is only once an hour.....
Well spotted Jeremy.
Dan, IMHO you're entitled to feel cheated. But as they say in all exams, you must read the question carefully. I'm sure Matthew will apologise....
Steve??????
Cheers
Don
Posted on: 28 April 2003 by Paul Ranson
quote:
The wheel rotates three times per hour
I thought it was four.
Three times from rubbing along the edges and once because the edges end up in a loop.
But probably not.
Paul
Posted on: 29 April 2003 by Matthew T
Nope, still wrong.
Matthew
Matthew
Posted on: 29 April 2003 by steved
Matthew,
Sorry about the duff maths earlier. What I meant to say is that the outside wheel will do 6 revolutions in 1 hour.
Try rolling a £1 coin round another £1 coin. The outside one actually does 2 revolutions.
Steve D
Sorry about the duff maths earlier. What I meant to say is that the outside wheel will do 6 revolutions in 1 hour.
Try rolling a £1 coin round another £1 coin. The outside one actually does 2 revolutions.
Steve D
Posted on: 30 April 2003 by Matthew T
Steve,
Thats the correct answer.
Matthew
Thats the correct answer.
Matthew
Posted on: 30 April 2003 by Dan M
Phew! Glad that one is sorted.
Dan
Dan
Posted on: 30 April 2003 by Don Atkinson
I had this suspicion it looked a bit like a cuckoo clock.....
nicely sorted steved
Cheers
Don
nicely sorted steved
Cheers
Don
Posted on: 02 May 2003 by Don Atkinson
Roots
This map shows the road layout where Janet and John live. How many different routes are there for them to cycle from home to school (together), never going northwards ?
Cheers
Don
This map shows the road layout where Janet and John live. How many different routes are there for them to cycle from home to school (together), never going northwards ?
Cheers
Don
Posted on: 02 May 2003 by JeremyD
Roots
First, label the successive nodes leading southwards between home and school a to h.
Then list the nodes reached by never-northward arcs from these nodes:
a -> b, c, e
b -> c, d
c -> d
d -> e, h
e -> f, g
f -> g, h
g -> h
Let n(xy) be the number of never-northward routes from x to y.
Then, using the above list:
n(gh) = 1
n(fh) = n(gh) + 1 = 2
n(eh) = n(fh) + n(gh) = 2 + 1 = 3
n(dh) = n(eh) + 1 = 3 + 1 = 4
n(ch) = n(dh) = 4
n(bh) = n(ch) + n(dh) = 4 + 4 = 8
n(ah) = n(bh) + n(ch) + n(eh) = 8 + 4 + 3 = 15
--J
First, label the successive nodes leading southwards between home and school a to h.
Then list the nodes reached by never-northward arcs from these nodes:
a -> b, c, e
b -> c, d
c -> d
d -> e, h
e -> f, g
f -> g, h
g -> h
Let n(xy) be the number of never-northward routes from x to y.
Then, using the above list:
n(gh) = 1
n(fh) = n(gh) + 1 = 2
n(eh) = n(fh) + n(gh) = 2 + 1 = 3
n(dh) = n(eh) + 1 = 3 + 1 = 4
n(ch) = n(dh) = 4
n(bh) = n(ch) + n(dh) = 4 + 4 = 8
n(ah) = n(bh) + n(ch) + n(eh) = 8 + 4 + 3 = 15
--J
Posted on: 05 May 2003 by Don Atkinson
JerremyD,
You're absolutely right and quite a neat explanation.
I thought I'd wait a while to see if there were any challengers, but none!
My solution ismuch more pedestrian (dreadful pun)
I labled allthe nodes, even those with just one route in and one route out.
Then I drew a tree (bit like a fault tree or probability tree) which branched whenever there was more than one route out.
The diagrams below should help
Cheers
Don
You're absolutely right and quite a neat explanation.
I thought I'd wait a while to see if there were any challengers, but none!
My solution ismuch more pedestrian (dreadful pun)
I labled allthe nodes, even those with just one route in and one route out.
Then I drew a tree (bit like a fault tree or probability tree) which branched whenever there was more than one route out.
The diagrams below should help
Cheers
Don
Posted on: 05 May 2003 by Don Atkinson
And the tree diagram
Cheers
Don
Cheers
Don