bonanova

I owe it to the Den to post at least one of these without error.
Possibly this one does that.

Give the longest route (sequence of city numbers) that visits all the cities
(a) not returning to starting city (sum of 7 distances - starting point matters)
(b) returning to starting city (sum of 8 distances - starting point does not matter)

This puzzle has more choices than the first one.

Cities lie clockwise on the perimeter of a 6x6 square:

6--O------O-------+--------O
   |3     4               5|
   |                       |
   |                       |
4--+                       +
   |                       |
   |                       |
   |                       |
2--O2                     6O
   |                       |
   |                       |
   |1              8      7|
0--O-------+-------O-------O
   |       |       |       |
   0       2       4       6


+----+---+---+
|City| x | y |   Distances:
+----+---+---+   8.485  1-5 3-7
| 1  | 0 | 0 |   7.211  2-5 3-6 3-8 4-7
| 2  | 0 | 2 |   6.325  1-4 1-6 2-7 4-8 5-8
| 3  | 0 | 6 |   6.000  1-3 1-7 2-6 3-5 5-7
| 4  | 2 | 6 |   5.656  4-6
| 5  | 6 | 6 |   4.472  2-4 2-8
| 6  | 6 | 2 |   4.000  1-8 2-3 4-5 5-6
| 7  | 6 | 0 |   2.828  6-8
| 8  | 4 | 0 |   2.000  1-2 3-4 6-7 7-8
+----+---+---+

Sergeant McGuffy remarked to his lieutenant one day after it was all over. "The way I reconstruct the crime, the innocent people were so excited when they first talked to us, they each got one fact wrong. But, the killer wanted to confuse our trail - and so coolly lied from beginning to end." Let's recap what we know ...

Egmont VanDorn has been found dead in his apartment. From the beginning, it is pretty clear to the police that this was no accident. Before they can be separated, the four young people who found the body - Arnold, Betsy, Charles and Daisy - eagerly begin to tell their story.
 

1. "If, as the police say, Egmont was injured between four and five," said Daisy, "that must let us all out. We were all together having dinner for at least two hours before we came over here."
    
2. "But we all arrived at seven," Charles pointed out.
    
3. "Arnold said that it was six o'clock at the time, just before we opened the door. Didn't you Arnold?" said Daisy.
    
4. "I said it was just seven on the nose," said Arnold. "Sorry, honey."
    
5. "I have no idea what he said or what time we got here," declared Betsy. "The thing I remember is the gas in the hall. We rang the doorbell, and no one came and no one came, and then I smelled the gas. My heart turned over, I thought to myself, he's dead. I just know he's dead."
    
6. "Don't dramatize yourself," said Daisy coolly. "There is no way that you could have smelled the gas before we opened the door. The place was locked up, and sealed too, tighter than a drum. We'd still be in the hall if Eggy hadn't given me a key last week."
    
7. "There was gas in the hall, all right," said Charles. "I smelled it before we opened the door. You seemed to take forever getting your key out. When you finally got the door opened, the gas just streamed out."
    
8. "That was a pretty dangerous thing you did, Charles," said Arnold, "turning on the lights the way you did. Didn't it occur to you that a spark at the light switch could have blown us all up."
    
9. "The lights were already on, Arnold," Charles replied.
    
10. "For my part, I'm sorry about pulling him out of the oven - tampering with the evidence and all that," said Arnold. "Murder never crossed my mind. Locked room, you know. All I could think of was that maybe he was alive and we could still save him."
     
11. "I don't believe this," said Daisy. "You didn't pull him out of the oven. I did. You ran and opened the window. Very good move, too, I thought at the time."
     
12. "I opened the window!" cried Betsy. "I was dizzy from the fumes, and I knew I needed to do something fast."
     
13. "The only things you opened were the door to the liquor cabinet and a bottle of Scotch, Betsy." Charles laughed at her. "And I thought they were good moves."
     
14. "That was Arnold who got out the Scotch," said Betsy. "Don't you remember our sitting there after you called the police, and Arnold passing out the glasses?"
     
15. "It couldn't have been me. Must have been you," said Arnold. "I've never been here before, I didn't even know where he kept the stuff. Charles, what did you do?"
     
16. "Do you know what? I don't think I did anything. I remember quite clearly, walking over here with you after that long dinner we had together, and then seeing poor Eggy's feet through the door. But after that, I don't think I did a thing, except stand there gasping."
     

OK it's time to make an arrest.

Which liar done it?

What if there were no diagonals?

 

1 2 3 4 5 6 7 8 9

Arrange these numbers into two separate groups so that they add up to same total.

Note : you cant turn 9 upside down and make it 6

 

Spoiler for Looks like

You have to turn the 9 into a six.

Numbers 1-9 total 45, which can't be evenly split.
Making the 9 a 6 reduces that total by 3 to 42.
So we need two groups that each total 21.

 

1 2 3 4 5 6 and 6 7 8 should do it.

 

Or turn the 6 into 9, increasing the total by 3 to 48.

 

1 2 3 4 5 9 and 7 8 9 do it for that case.

this is what even i think how can you split 45 into two equal parts...P.S- he noted that we cant use 9 as 6...

Ah, I misread that. Good catch.

Spoiler for

31 and 10000

Spoiler for explanation

Number 16 represented in base 16-n, for n in 1..14

  Good solve ...  although I got the "basic" nature of the puzzle, I did not see the second "?" in the OP.

 

Suppose n tickets numbered 1-n are in a box, and you draw one of them
You don't know how many tickets are in the box, but you are asked to estimate how many there are.
Your ticket has the number p on it.
 
What estimate of n has the highest likelihood of being correct?

 
It seems like this problem is dependent upon what is a reasonable a priori distribution for N.

Spoiler for

Let P(N) be the prior distribution for the number N. Let P(p|N) be the conditional probability for p given N; we know that P(p|N) = 1/N. We wish to compute the conditional probability P(N|p), which can be expanded using Bayes theorem
bayesian.png
 
So, if we are given a well-defined probability function P(N), we can easily compute the number N that gives that maximum probability P(N|p).
 
The problem is that we are not given what P(N) is. One reasonable choice would be to assume that P(N) is the improper uniform distribution from p to infinity (every number has an equal chance of occuring). However, if we plug that improper distribution into the equation above, we would be able to cancel out P(N), but we'd end up with a harmonic series in the denominator, which does not converge.
 

 
 
Oh wait, the answer is much simpler than that. Turns out this is the easiest bonanova puzzle I have ever seen =)

Spoiler for

Since "n tickets numbered 1-n are in a box", if we get number p = (1-n), then it is easy to see that n = 1 - p.

LOL

That may be the first and only time I use LOL in this forum.

I'm not that clever. Really.

But I love it. It's a better puzzle than the one I intended.

Honorable mention.

.

Spoiler for is Russian translation available?

If "...the weight of the weight and the weight of the rope were together half as much again as the weight of the monkey," how can that system be in equilibrium?

 Конечно ...   Веревка подбежал шкив, на одном конце была обезьяна, на другом конце вес.Два оставшихся в равновесии.Вес каната было 4 унции / футбол, а в возрасте от обезьяны и мать обезьяны составил четыре года.Вес обезьяна была много фунтов, как мать обезьяны было лет, и весом весом и весом каната вместе были в полтора раза больше, поскольку вес обезьян.
 
Вес вес превышал вес веревки, как многие фунтов, как обезьяна лет, когда мать обезьяны был вдвое старше брата обезьяны было, когда мать обезьяны был вдвое старше брата обезьяна будет, когда брат обезьяны в три раза стара, как мать обезьяны Когда мама была обезьяна была в три раза стара, как обезьяна в пункте 1.
 
Мать обезьяны был вдвое старше обезьяна Когда мать обезьяны был наполовину стара, как обезьяна будет Когда обезьяна в три раза стара, как мать обезьяны было, когда мать обезьяны в три раза стара, как обезьяна в пункте 1.
 
Возраст матери обезьяны превысил возраст брата обезьяны на такую ​​же сумму, как возраст брата обезьяны превысил возраст обезьяны.
 
Какова была длина веревки?

       So the rope has legs and plays footbal?! And monkey's mother is a male?! And age is money?!Was it Google translation? I guess, it'll be awhile before computers can understand and translate natural language.Although it's worth noting that about the only bit translated correctly, is that very thing, which I misunderstood. The translation states, the weight of the rope together with the weight is 1.5 times the weight of the monkey, as the problem statement intended. Whereas I understood it as half weight of the monkey.

Spoiler for what about this

AS QC JH KD

JD KS AC QH

KC JS QD AH

QH AD KS JC

IN SECOND ROW  SECOND COLUMN IT SHOULD BE 'KH'. I TRIED TO EDIT THE POST BUT EDITING FAILED.

Hint: To edit after time elapses, copy / edit / paste onto a new post.

Here's what I think.

I am giving this puzzle verbatim from a book I read long ago. These days most everything I have ever done was long ago. But I digress. I simply don't trust myself to re-cast it into scintillating banter style. So if its tenor is archaic or arcane, the onus is off me.  Enjoy.
 
A rope ran over a pulley; at one end was a monkey, at the other end a weight. The two remained in equilibrium. The weight of the rope was 4 oz/foot, and the ages of the monkey and the monkey's mother amounted to four years. The weight of the monkey was as many pounds as the monkey's mother was years old, and the weight of the weight and the weight of the rope were together half as much again as the weight of the monkey.
 
The weight of the weight exceeded the weight of the rope by as many pounds as the monkey was years old when the monkey's mother was twice as old as the monkey's brother was when the monkey's mother was twice half as old as the monkey's brother will be when the monkey's brother is three times as old as the monkey's mother was when the monkey's mother was three times as old as the monkey was in paragraph 1.
 
The monkey's mother was twice as old as the monkey was when the monkey's mother was half as old as the monkey will be when the monkey is three times as old as the monkey's mother was when the monkey's mother was three times as old as the monkey was in paragraph 1.
 
The age of the monkey's mother exceeded the age of the monkey's brother by the same amount as the age of the monkey's brother exceeded the age of the monkey.
 
What was the length of the rope?

Professor bonanova peeks in, wondering if there is a clue yet.

He certainly has none ...

I agree. Which side of the debate would you take?