Puzzles with numbers and things

[berolina]

Final call for a 5th answer. In any case, I'll close this round in 10 minutes....

Nooooo!

[davis_greatest]

Final call for a 5th answer. In any case, I'll close this round in 10 minutes....

http://www.youtube.com/watch?v=zxWK9keh3XI

Round 121 - When I'm cleaning windas... di diddleedeedee di diddleedeedee di diddleedeedee ---- DEE DEE!

Forge Geormby goes and asks Scrooge McDott whether he would like his windows cleaned. Scrooge's 10 windows, each measuring 2 feet by 5 feet, have not been cleaned for 3 years, so Scrooge says he will have them cleaned but will pay Forge a penny for every five square feet of window washed, and not a penny more.

Forge is not very good at arithmetic, so rather than trying to figure out how much Scrooge has offered, he said that he would wash the windows for 30 pence.

Scrooge has a think. Does he employ Forge's services?

Answers.... on this thread!
Please just answer Yes or No, without any explanation. I will wait till at least 5 people have answered (hopefully that won't take forever) before posting the answer. So you can still answer if someone else has.

[davis_greatest]

Would you please tell us where you got this nice formula?

There are 3 possibilities:

(A) breaks of k reds with colours, followed by one red. Number of such breaks is 6^k (since there are 6 possible colours with each of the k reds).
Add 6^k for k=0,1,2,...,14 and you get (6^15-1)/(6-1) = (6^15-1)/5

(B) breaks of k reds with colours, followed by anything from 0 to 6 of the final colours (0 would be no final colours, 6 would be yellow, green, brown, blue, pink, black).
Number of such breaks is (6^k) x 7 (since potting between 0 and 6 final colours gives 7 possibilities)
Add (6^k) x 7 for k=1,2,3,...,15 and you get 6 x (A) x 7 = 42 (A).

(C) Potting 1 of the 6 final colours: 6 ways (yellow or green or brown or blue or pink or black)
Potting 2 of the 6 final colours: 5 ways
Potting 3 of the 6 final colours: 4 ways
Potting 4 of the 6 final colours: 3 ways
Potting 5 of the 6 final colours: 2 ways
Potting all of the 6 final colours: 1 way

6+5+4+3+2+1 = 21


So total is (A) + 42(A) + 21 = 43(A) + 21 = 43/5x(6^15-1) + 21



If we had r reds and c colours, there would be

(c^2 + c + 1)/(c-1) x (c^r-1) + c(c+1) / 2
ways of making breaks.

Normal snooker has r=15, c=6.

[abextra]

Congratulations to snookersfun who correctly counted there being 4,043,590,867,366 possible breaks. In fact, the easiest way (at least, my way) to obtain this number is to see that there are 43/5*(6^15-1) + 21 possible breaks! (Explanation available on request.)

Would you please tell us where you got this nice formula?

[Snooker Rocks!]

OK can I do a question like snookersfun's?

THE CHILD WITH THE WART:

A: "What are the ages, in years only, of your three children?"
B: "The product of their ages of 36."
A: "Not enough information."
B: "The sum of their ages equals your house number."
A: "Still not enough information."
B: "My oldest son - and he's at least a year older than either of the others - has a wart on his left thumb."
A: "That's enough, thank you. Their ages are...."

Complete A's sentence.

If no one has done it by 20:00 tomorrow, I'll reveal the answer

I had forgotten about this question, but a very kind positive rep by d_g for questions on this thread reminded me that I never answered this one! Thankyou

Ok, very well done to: davis_greatest, snookersfun and abextra for getting the answer to this question right. The ages were 2, 2, 9.

After this line,

B: "The product of their ages of 36."

You know that their ages are one of these, because these are the only 8 triplets that have a product of 36:

1, 1, 36
1, 2, 18
1, 3, 12
1, 4, 9
1, 6, 6
2, 2, 9
2, 3, 6
3, 4, 3

After this line,

B: "The sum of their ages equals your house number."

You can erase all triplets which have a different sum (because speaker A would have known his house number), so you are left with:

1, 1, 36
1, 2, 18
1, 3, 12
1, 4, 9
1, 6, 6
2, 2, 9
2, 3, 6
3, 4, 3

After this line,

B: "My oldest son - and he's at least a year older than either of the others - has a wart on his left thumb."

You can erase 1, 6, 6, because you know that there is an oldest son, older by at least one year.

You are therefore left with:

1, 1, 36
1, 2, 18
1, 3, 12
1, 4, 9
1, 6, 6
2, 2, 9
2, 3, 6
3, 4, 3

And 2, 2, 9 was the correct answer!

[davis_greatest]

We've had 2 x Yes, 2 x No, and one "Price is not important".

Who wants to cast the deciding vote for the viewers?

[dantuck_7]

Price is not important - its about the quality of the service offered.

[snookersfun]

no

[The Statman]

Yes, definitely

[April madness]

Round 121: somehow I think Scrooge will say No to such an expensive window cleaning service. Hopefully I am not completely wrong

[abextra]

Yes?

[davis_greatest]

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?

I've had two correct answers - from snookersfun and abextra. Congratulations!

The next person to answer round 120 above, please answer on this thread.


At the same time, here is an easy one...

Round 121 - When I'm cleaning windas... di diddleedeedee di diddleedeedee di diddleedeedee ---- DEE DEE!

Forge Geormby goes and asks Scrooge McDott whether he would like his windows cleaned. Scrooge's 10 windows, each measuring 2 feet by 5 feet, have not been cleaned for 3 years, so Scrooge says he will have them cleaned but will pay Forge a penny for every five square feet of window washed, and not a penny more.

Forge is not very good at arithmetic, so rather than trying to figure out how much Scrooge has offered, he said that he would wash the windows for 30 pence.

Scrooge has a think. Does he employ Forge's services?

Answers.... on this thread!
Please just answer Yes or No, without any explanation. I will wait till at least 5 people have answered (hopefully that won't take forever) before posting the answer. So you can still answer if someone else has.

[davis_greatest]

Can you clarify whether you would get a half point ofor winning the last match? Or is this not counted?

Dan.

The winner of the final frame scores one frame for that frame.

[dantuck_7]

Can you clarify whether you would get a half point ofor winning the last match? Or is this not counted?

Dan.

[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?