An Assortment Of Puzzles

Originally posted by THUDandBLUNDER
Any lawyer would be happy to sniff out double meanings etc.
At the right price, of course.

I don't care what price a crab-sniffing lawyer charges, I ain't payin' it!

Originally posted by phgao
Here's another

There are n prisoners in a row of locked cells. With the return of the King from the Crusades, a partial amnesty is declared and it works like this. When the prisoners are still asleap, the jailer walks past the cells n tim ...[text shortened]... cked at the ned of the process so that the prisoner is free to go?

All cells corresponding to perfect squares are open! Why you ask? Well, it has to do with the number of factors for each cell number.

If the number of key turns for a given cell is odd, the cell is open. If it's even, the cell remains closed. In this problem the lock will be toggled open/closed if the cell number is divisible by the pass, for example 12 will be toggled on passes 1, 2, 3, 4, 6 and 12 (all the factors of 12).

Now, all the factors come in pairs, for example 12 = 1*12 = 2*6 = 3*4, which makes an even number of factors, leaving the cell door locked. Perfect square factors also come in pairs, but with a twist: the square root is used twice in one factoring, for example 9 = 3*3. Since you only toggle the lock once on this pass, the total number of toggles is odd, leaving the cell door open!

This also makes the king a bit of a d*ck, as the number of cells open will be no higher than n^(1/2). For example, with 1,000 prisoners only 31 (3.1% ) will go free; with 10,000 prisoners, only 100 (1% ) will go free. And Lord knows, the serial killers will probably get out first and the guy who stole a piece of mutton will still be locked up. He should be in college! This is what's wrong with the justice system.

Originally posted by PBE6
Technically, the terms "husband" and "wife" apply to married homosexuals, too. And I've watched enough "L Word" and "Queer As Folk" that I should have snarked out Nordlys' post myself.

It should be 'husband and husband' or 'wife and wife' for homosexuals. That would be more accurate, would it not?

Originally posted by BigDoggProblem
It should be 'husband and husband' or 'wife and wife' for homosexuals. That would be more accurate, would it not?

Depends which one plays with the LPGA.

🙄!!!

Originally posted by phgao
Plus the 'couple' thing is explained in the Q- ...no woman shakes hands with either her husband...!! So why the confusion?

Because "no woman shakes hands etc. etc." doesn't say anything about how many women there are. I think you are the one who is confused. 😉

This is getting out of hand.

Nordlys, your observation was (as I'm sure you'll agree) a bit anal, but was also an interesting wrinkle on the problem and a zany satire on contemporary mores. Zingo for you. THUD, your response was (as I'm sure you'll agree) a bit whiny, but made common sense nonetheless. Zingo for you. The problem was solved as intended, everyone looks smart, zingos all around.

My comment was posted mainly so I could say "dead below the waist". Ba-ZING!!!

Now let's all put it back in our trousers and get on with the business of problem solving.

Here's one I hope no one has heard before (although with this bunch, that's less than likely):

A parking lane 4 units long is nestled in the heart of Busy Town's busy downtown. Unfortunately, no parking spaces have been delineated by city staff. Being an ultra sweet spot, cars of unit length are always trying to park there. Being concientuous, each driver parks perfectly parallel and tightly adjacent to the sidewalk. However, not being very courteous, they don't care where along the length of the lane they park. On average, how many cars will be able to park at once on the lane? (Assume that every driver is a cheap pr;ck, and won't risk a ticket by having part of their car hang over the edge of the lane. Also assume that every driver is so adept at parallel parking it's scary.)

Originally posted by phgao
One more thing b4 I go,

The answer to Q2, is 120 degrees not 90 degrees, note: the largest possible size of an angle of a triangle in a Square is 90 degrees!

If the vertices of the cube have coordinates
(0,0,0)
(0,0,1)
(0,1,0)
(1,0,0)
(1,1,0)
(1,0,1)
(0,1,1)
(1,1,1)
please give the coordinates of 3 such mid-points.

Ok, the angle size is 120 degrees. See if you can see how.

With the vertices of the cube given by

(0,0,0), (0,0,2), (0,2,2), (0,2,0),
(2,0,0), (2,0,2), (2,2,2), (2,2,0)

Consider the triangle formed by
(0,0,1), (0,1,0), (1,2,0)

The cosine of the ange at (0,1,0) is
cos(a) = ((0,0,1)-(0,1,0)) o ((1,2,0)-(0,1,0)) / m
m = sqrt(|0,0,1)-(0,1,0)|^2 x |(1,2,0)-(0,1,0)|^2)
= 2

=> cos(a) = -1/2 => a = 120 deg

Originally posted by phgao
With the vertices of the cube given by

(0,0,0), (0,0,2), (0,2,2), (0,2,0),
(2,0,0), (2,0,2), (2,2,2), (2,2,0)

Consider the triangle formed by
(0,0,1), (0,1,0), (1,2,0)

The cosine of the ange at (0,1,0) is
cos(a) = ((0,0,1)-(0,1,0)) o ((1,2,0)-(0,1,0)) / m
m = sqrt(|0,0,1)-(0,1,0)|^2 x |(1,2,0)-(0,1,0)|^2)
= 2

=> cos(a) = -1/2 => a = 120 deg

THUDandBLUNDER is that right?