Brain Teasers

[anj]

A puzzle in honor of Hadaso -- Happy Birthday!

At one point, a remote island's population of chameleons was divided as follows:

* 13 red chameleons
* 15 green chameleons
* 17 blue chameleons

Each time two different colored chameleons would meet, they would change their color to the third one. (i.e.. If green meets red, they both change their color to blue.) Is it ever possible for all chameleons to become the same color? Why or why not?"

[CyberSmurf]

No, because there can never be an equal number of chameleons for any two colours.

If a green and blue chameleon meet the new counts will be ...
... 14 green (15-1)
... 15 red (13+2)
... 16 blue (17-1)
It this exchange red has moved up a net +3 in relation to each of the other two. But the difference between green and blue did not change.

...

However, if you stepped on one of the blue chameleons ... ...

[newhampshire]

Also interesting, that the green animals suffer. In every simulation I've done the green numbers stay low.

A more in depth question for hadaso to answer.

See runs below...

http://animals.tempweb.mm.st/

[hadaso]

If one divides the three numbers by 3 the remainders are 0,1,2. If two chameleons meet the number of chameleons of each color changes but the remainders remain 0,1,2 (in different order). So it's not possible to have the same number of chameleons of two different colors.

A similar problem ("stolen" from the math olympics for grades 7-9, stage 1, Weizmann Institute of science, this year. It was the most difficult problem there, I think):

Consider an infinite chessboard. It is not difficult to see that a knight that is in some square can reach any other square. The same is not true for a bishop (a bishop on a white square can only reach white squares). Now we introduce a new piece: "the mxn chariot". This piece always move m squares in one direction and n squares in the other direction. For instance: a knight is a 1x2 chariot. A bishop is not a chariot but a 1x1 chariot can reach exactly the same squares as a bishop does (using more turns. A 1x1 chariot can move 1 square diagonaly in each turn).

The problem is to determine which chariots can reach every squre on the board (actually this was part three of the problem. The first two part asked about particular chariots and were much more appropriate for the kids. IMO the general problem was a bit too much for them but right now I don't remember the exact numbers they asked about.)

[CyberSmurf]

Hmm, I don't have the higher math skill for a concise answer on that one.

But assuming that neither m or n can be 0, and n is always greater or equal to m ...

If n is a multiple of m then the piece may only land on squares defined by an m by m grid.
... In the case of a knight this would be a 1 by 1 grid.
... For a 4x2 or 6x2 chariot this would be a 2 by 2 grid.
If m+n totals an even number, then the piece will always stay on the same colour.
If neither of those conditions are met, then the piece can move to any square, given enough moves.

Did I miss anything?

[anj]

Lovely! One more puzzle from the devious collection at http://techinterview.org. I'm picking the less verbose ones, but there are some fun scenario ones too.

part I: what is the angle between the minute hand and the hour hand at 3:15 on an analog clock? no, its not 0.

part II: how often does the minute hand pass the hour hand on an analog clock?

[CyberSmurf]

1. 7½º ... (postion of hour hand minus minute hand = 3.25÷12×360º - 90º)
   
   
2. In a 24 hour period the minute hand makes 24 revolutions and the hour hand makes 2 revolutions. So the minute hand pases the hour hand 22 times a day.

[hadaso]

Originally Posted by CyberSmurf

Hmm, I don't have the higher math skill for a concise answer on that one.

But assuming that neither m or n can be 0, and n is always greater or equal to m ...

If n is a multiple of m then the piece may only land on squares defined by an m by m grid.
... In the case of a knight this would be a 1 by 1 grid.
... For a 4x2 or 6x2 chariot this would be a 2 by 2 grid.
If m+n totals an even number, then the piece will always stay on the same colour.
If neither of those conditions are met, then the piece can move to any square, given enough moves.

Did I miss anything?

Not really, except that also if both m and n are multiple of the same number then the chariot is also limited, for instance a 6x9 chariot is limited to a 3 by 3 grid. The more difficult part is showing why in all other cases (m,n have no common divisor except 1 and m+n is odd) every square can actually be reached. For instance why a 1x8 or 7x18 chariot actually can get to any square.

[anj]

Originally Posted by CyberSmurf

In a 24 hour period the minute hand makes 24 revolutions and the hour hand makes 2 revolutions. So the minute hand pases the hour hand 22 times a day.

Well this is certainly the most elegant way of putting it! It's almost the same, but I think they are looking to know how much time passes between two meetings of the hour and the minute hands.

[CyberSmurf]

Would that be 1 hour, 5 minutes, 27 and 3/11 seconds then? ...

[anj]

Yup! The answer at the site where I got the puzzle gets around the problem by saying "every 1 1/11 hours". Of course, hands on clocks don't advance unstintingly (like sand through an hourglass) but rather in increments.

It is difficult to answer exactly without knowing more about the mechanism by which the clock advances. The hands advance incrementally at some interval, right? But what are those increments and intervals?

When I first read the puzzle, I somehow assumed the hands advance in 6 degree increments, such that the minute hand advances at whole minute intervals (6 degrees/ minute) and the hour hand advances throughout the hour by 6 degrees every 12 minutes. I think I just assumed that from observing the second hand, that there are 5 "steps" between each of the 12 markings. If that were true, the hands would line up 5 minutes after the "first" hour and the "second hour", and 6 minutes after the third hour.

However in looking at a watch, I realized that the minute hand can actually advance incrementally throughout the minute, as well. That increment is determined by the mechanics of the clock, so in a mechanical clock it actually depends on the number of teeth on each gear and the gear ratios.

On that subject, these animations of clock escapements here are interesting.

[robert@fm]

Given the following word list:

AN, AS, ER, OR, NOT, PEN, SET, SOP, ...

1) What is the rationale behind this list?

2) What ninth word completes the list?

Clue:  I've posed this one before in a very different form.

(edit: there are actually at least four possible "ninth words", the order of the letters doesn't matter...)

[robert@fm]

Word list (the last post before this one -- hint, think in groups of three)

'39 -- meaning of lyrics, 22nd September 2006 (requires lateral thinking)

Song odd-one-out, 18th September 2006 (that one is one short of "evil")

The Appointment, 21st November 2005 (hinges on British English, we don't say times like "ten of five" here, we say "ten to five" meaning ten minutes until the hour)

Fiscal odd-one-out, 7th July 2005

Round Dollar Amount, 5th November 2002 -- solved?

Moon odd-one-out, 8th September 2004

The Hotel, 30th July 2004 (although I received a few solutions, all correct, by PM)

DitLoIDs, 17th June 2004 (all but two)


Previous unsolved-puzzles roundup, 9th April 2004

[hadaso]

Originally Posted by robert@fm

Round Dollar Amount, 5th November 2002 -- solved?

Aha! (this means yes - I need at least 10 chars to post!)

[Mal]

Came back to this extensive thread to find the answer to a puzzle I thought about for a long time - the one about the ABCBA advertising boards, and visited the unsolved ones.

The solution to the The Hotel may just have occurred to me.

Playing Monopoly

Highlight the previous line to reveal my idea.

== Mal