The ‘basic’ teaser is illustrated in the diagram below.
A rope passes over a pulley, which in turn is fixed to the roof of a building. At one end of the rope a weight is fastened, which exactly counterbalances a monkey which is hanging on the other end of the rope.
If the monkey starts to climb the rope, what will be the result ?
PS The pulley is perfectly frictionless, and the rope is perfectly light and flexible.
I’ll have a Christmas stab and say nothing. 
I have my own assessment of what the result will be if the monkey starts to climb the rope, however … I have been unable to verify it despite an extensive web-search.
A few web-sites outline a result that is the same, or very similar to my result. However, I haven’t found a reliable site, with a peer-reviewed outcome and verified solution.
I believe that the initial teaser might have been set by a certain Mr Charles Dodgson, an Oxford University mathematician, better known as … ?
I’m not very familiar with his work but it does sound like the sort of conundrum that he might use in his stories.
I have an answer but it is largely based on basic principles of physics (mechanics) that I last studied in 1978, plus a bit of gut feel. So I could be wrong.
I will post my thoughts tomorrow, if others want to have a stab first.
Meanwhile, here’s this year’s GCHQ Challenge. Getting all the clues is always a bit tricky but worth having a go. As usual I guessed a couple of them rather than work them out formally.
I was thinking more along the lines of Xmas Hymns, Xmas Songs, Xmas … you know ! Hark the Herald Angels, Silent Night … that sort of music.
Wow, I’d forgotten about that, Ravvie. Thanks for reminding us all.
Cheers
Don
As long as said Money is climbing, he will move up the rope, but as soon as he stops he will lower again to the same height as the weight.
Hi Mike,
Your outcome is the same as my outcome ie when the monkey stops climbing, the weight and the monkey are in the same relative positions as when they started, and both are closer to the pulley.
But I don’t think that the monkey climbs up (say) 8 ft without the rope moving and only then is followed by the rope moving round the pulley, for the weight to rise (say) 4ft and the monkey to lower by 4ft back to their relative starting positions but 4ft closer to the pulley.
For the sake of simplicity, we should assume that the rope is ‘light’ ie ignore its weight. I have only just realised I should have stated this in the initial teaser. I amended the teaser on 24/12/2025.
When the monkey pulls on the rope it is exerting a force. This increases the tension in the rope beyond the weight of the monkey or counterweight.
The tension applies the same force to the counterweight (principle of equal and opposite reaction).
Hence both the monkey and the counterweight will rise equally.
When the monkey stops pulling, both monkey and counterweight will remain in their new positions.
So that we can all relax over the extended New Year holiday weekend, I felt it might be thoughtful if I provided the answers to the above two teasers.
Anyone giving one or other teaser a try, would know whether to relax with a glass of well-earned whisky, or drown their sorrows with a bottle of Gin.
The diameters of the two circles are 50mm and 41mm
There are 32 triangles in the hexagonal shaped puzzle.
Enjoy
Don
The triangle/hexagon puzzle suited me due to being able to visualise the regular patterns:
6x2 small triangles
6x2 pairs of small triangles
2 large triangles
6 triples along each side of the large triangles.
Mrs R chose to count them, colouring particular types. It took somewhat longer but we independently reached 32.
For the “moon crescent” puzzle, my initial answer felt a bit clunky, so I asked Mrs R if she knew of a simple way (she teaches geometry amongst other topics). She agreed the answer but it was actually a bit more involved than my first effort.
I then ended up with this:
If the centre of the small circle is M then use Pythagoras on triangle FDM:
FM = radius of small circle = r, say.
DM = 4.5
ED = radius of large circle = r + 4.5
FD = r + 4.5 - 5 = r - 0.5
r^2 = (r - 0.5)^2 + 4.5^2 using Pythagoras
r = 0.5^2 + 4.5^2 = 20.5
Giving 41mm and 50mm
Nice solutions Ravvie.
For the triangles, I used the same technique as Mrs D.
For the crescent, I used the Intersecting Chords Theorem - i’m sure someone in addition to myself can recall that theorem ?
Has anyone made any progress on the GCHQ Christmas Challenge?
A game of chess, or a game of chance ?
A short while ago we spent a few weeks with my daughter and her family in BC Canada. We enjoyed many days outdoors hiking and canoeing. In the evenings, my son-in-law was keen to play cards, which we did. He’s fairly good but by the end of our holiday, even though our stakes were only in Cents rather than Dollars, he was in debt to me to the tune of $64. I suggested he might like a little fun using a standard chess board to clear his debt.
All he had to do was place a single Cent in the first square (top-left) of the chess board, then moving one square to the right, place 2 cents in the second square, 4 cents in the third square, 8 cents in the 4th square and so forth, zig-zagging across and down the board, doubling the number of cents at each new square, until all 64 squares were completed.
All the money in all 64 squares would be mine, and he would be clear of the debt, even if that total was less than $64.
Was he wise to take the challenge, or should he have settled up with $64 ?
He should have settled …
It would be more than the total world GDP.
I could give the number … or not, what would you prefer?
Nicely done WalnutTree, very nicely done !
You have the privilege of posting the number as and when it suits you. Well done.
