Puzzles with numbers and things

[berolina]

You should change the name of the thread to puzzles with numbers and apes...anyway my guess is 1556.

[rambon]

You should change the name of the thread to puzzles with numbers and apes...anyway my guess is 1556.

Sorry Berolina, and easy mistake to make but that's incorrect

[berolina]

Sorry Berolina, and easy mistake to make but that's incorrect

Well, I'm not used to playing snooker with apes...

[rambon]

Well, I'm not used to playing snooker with apes...

Don't worry about it... You were on the right lines, but you just need to think one step further.....

[berolina]

Don't worry about it... You were on the right lines, but you just need to think one step further.....

Monkeys?
I can't think right now...I guess I have to leave this thread in shame...my maths days are over...I'll go to the zoo tomorrow...my last guess is 492.

[davis_greatest]

Congratulations berolina!


So here, with new joiner berolina, is the latest Puzzles with numbers and things Hall of Frame

Oliver (my pet orang-utan)
Gordon (my pet gorilla)
Charlie (my pet chimpanzee)
snookersfun
abextra
davis_greatest (Oliver's, Gordon's and Charlie's pet something)
Vidas
chasmmi
elvaago
robert602
Sarmu
The Statman
austrian_girl
austrian_girl's dad
Semih_Sayginer
Snooker Rocks!
Ginger_Freak
April Madness
steveb72
rambon
Microsoft Excel
dantuck_7
berolina

[rambon]

492 is the correct answer, and there are a few ways of doing it.

We need 2048 people in the next round, which means we need to lose 3604-2048 apes, or 1556. This means we have to have 1556 matches, which will require 3112 apes.

Therefore we do not require 3604 -3112 apes, or 492

ALTERNATIVE SOLUTION

In an ideal world, we would have 4096 apes (2 to power 12), which would mean we could have 2048, then 1024.

So we simply subtract 3604 from 4096 and get 492

[berolina]

:d :d :d :d

[davis_greatest]

ALTERNATIVE SOLUTION

In an ideal world, we would have 4096 apes (2 to power 12), which would mean we could have 2048, then 1024.

So we simply subtract 3604 from 4096 and get 492

This of course is the easier way. Congratulations also to April Madness and to snookersfun. rambon, did you get any other answers to be congratulated?

[davis_greatest]

Well, I'm not used to playing snooker with apes...

I am. In fact, this will be the subject of the next question (once I've posted my predictions and done a quick bit of urgent work).

[rambon]

This of course is the easier way. Congratulations also to April Madness and to snookersfun. rambon, did you get any other answers to be congratulated?

Apart from yourself davis_greatest, no!

A quick one for you all now. The final of the tournament, saw Chim Pan Zee (a primate from China) play Gor Illa (an Albanian primate). Every time a ball was potted, bizarrely, the value of the ball equalled the number of the next shots to be taken without potting a ball). Chim Pan Zee broke, and went in-off the pack, and Gor Illa potted a red with his first shot. Given that no more fouls occurred and no-one potted more than one ball at a time, who won the frame and by what score?

[rambon]

Congratulations to snookersfun who got the right answer (eventually!!!). Will leave it open for a while though to see if anyone else gets there

[rambon]

Berolina now has it right as well so I'l post the answer hidden here for those who want to check it.

Chim broke off and went in-off the pack, which meant that Gor lead 0-4
Gor potted a red (rules say the next shot was missed) 0-5
Chim potted a red (rules say the next shot was missed) 1-5
Reds are now alternately potted by each player (missing the colour each time) until the score reaches Chim 7 Gor 12
Chim has a go at the yellow first, which he pots to make it 9-12, but he misses the green (miss 1) and Gor misses the green (miss 2 = value of yellow). Chim therefore pots the green to make it 12-12
Chim, Gor and Chim (3 times = value of green) miss the brown before Gor pots it: 12-16
Gor, Chim, Gor and Chim (4 times = value of brown) miss the blue before Gor pots it: 12-21
Gor, Chim, Gor, Chim and Gor (5 times = value of blue) miss the pink before Chim pots it: 18-21
Chim, Gor, Chim, Gor, Chim and Gor (6 times = value of pink) miss the black before Chim pots it: 25-21

So the final score is Chim Pan Zee 25 Gor Illa 21

[davis_greatest]

Berolina now has it right as well so I'l post the answer hidden here for those who want to check it.

Well, I agreed that. I think you posted rounds 118 and 119, so I'll follow with round 120...

[davis_greatest]

Round 120 - You wait all day for an ape, and then they come in threes

Well, I'm not used to playing snooker with apes...

This, I guarantee you, is a (99%) true story. I have only changed the odd, very minor detail (if you think you know which detail I have changed, you can guess it, and may earn some more fame, but that is not the point of this question)...

Round 120 - You wait all day for an ape, and then they come in threes

Last night, I played snooker with my apes. Usually, there are four of us: me, my pet orang-utan Oliver, my pet gorilla Gordon and, of course, my pet chimpanzee Charlie. Occasionally, one cannot make it, and last night was one of those nights.

When there are three of us playing, we play one against two (a bit like doubles, but the "one" on his own also plays the shots for his imaginary doubles partner). Then we rotate, so that we each play one frame alone.

Last night, with my opening visit, I played a very nice indeed long red, with a lot of left hand side to avoid the cue ball cannoning into reds after hitting the top cushion (as the black was tied up), thus finishing nicely on the blue. Potting the blue at pace enabled a full ball cannon onto the pink, to split the reds, one of which I then potted, followed by a pink... but I digress....

Depending on the standard, and the amount of safety played, sometimes we have a bit more time and play a 4th frame (sometimes with a smaller number of reds, such as 10). And so it was last night. For the 4th frame, all three of us play against each other, i.e. three "teams" of one, so we each come to the table every 3rd visit. This entirely ruins the tactical aspect, but it's still fun enough at the end of the evening...

If we counted the frames, a fair way to score it might be to award one frame to each person / ape for each frame won, except for frames won by a pair, in which case each member of the pair would be awarded half a frame.

Scoring the frames that way last night would have meant that everyone scored, we all got different scores, and I, of course, got the highest.

What was my outcome (win/loss) for each frame?