Brain Teaser No 1

Hello all,

I'm reading a fictional book a friend gave me
on code breakers set (in part) in WWII. One
scene has the following set of numbers
written on a chalk board:


19 17 17 19 14 20 23 18 19 8 12 16 19 8 3

21 8 25 18 14 18 6 3 18 8 15 18 22 18 11


The book then goes on to show what phrase
this spells -- the method seemed
straighforward, but then again I had the
answer in front of me. Anyone care to decipher
it?

Dan

No takers? Here's a hint -- no number is larger than 26.

cheers

Dan

Dan,

Its Xmas.

In the UK we still have Public Holidays to celebrate mainstream Christian events (Xmas; Good Friday; Easter Monday - Whitsuntide got moved to a fixed date a few years back and effectively became secular)

Because of the closeness of Xmas and new year, half the nation seems to have opted for a few days annual leave and has taken a two-week break! Back to normal on the 5th January!!!

Also, we have Xmas-cracker jokes and puzzles to cope with......

and hang-overs.....

Don't despair, we'll get back to crack the code soon....

Merry Xmas

Don

Don,

Yes, you're right - being a Brit myself and having worked for firms in the UK I should have known most will be taking a good break -- something the US would do well to emulate.

I'll let this one sit a while...

cheers,

Dan

quote:

---

Originally posted by Dan M:
No takers? Here's a hint -- no number is larger than 26.

---

Another hint might have been "you don't need to dust off the enigma machine for this one".

ATTACK PEARL HARBOR DECEMBER SEVEN

Started by assuming that the first word was 4 or more letters and looked for a word whose first 4 letters fit the pattern ABBA that either had a war or Christmas bent. Once I found "ATTACK" and plugged in the other "A"s and "C"s and the "E"s (assuming that the most common number - 18 = "E") the answer was pretty obvious.

I knew those cryptic crosswords would come in handy one day.

Nicely done Minky!

"you don't need to dust off the enigma machine for this one".

I had always assumed that the enigma machine had to decode messages written in German (as opposed to English).

I'm not sure whether the Japaneese coded their messeges in Japaneese, English or what....

Cheers

Don

http://www.worldwar2history.info/Midway/Japanese.html and onwards is quite interesting.

Paul

An easy warm-up

Well, its been a while since we tried out the old grey cells, so a gentle warm-up seemed like a good idea....

I have an equalateral triangle and a regular hexagon, and their perimeters are of equal length.

What is the ratio of their areas?

Cheers

Don

The perimeter is P.

The triangle has sides length P/3.
The hexagon has sides length P/6.

A hexagon can be tiled with 6 equilateral triangles with sides P/6. Such a triangle has an area 1/4 of one with P/3 sides. So the hexagon covers 6/4 the area of the triangle.

Paul

Well done Paul.

There's nearly always lots of ways to skin a cat.....

When I first met this particular teaser, I admit that I jumpped straight in with the 'semi-perimeter' formula for the area of a triangle. Only took a couple of minutes and delivered the right answer. I didn't bother with a drawing at all.

Then, a few days later, whilst doodling, the drawing emmerged showing the 4 smaller triangles enclosed within the larger triangle.

I could only smile.

Cheers

Don

That's neater. I just saw that the lengths were half so the area was a quarter and jumped...

Paul

That's neater.

and here's the picture I had in mind

Cheers

Don

3D Topology

My cousin is a dustman.

He works in Salisbury and last week, when emptying a dustbin at a certain factory, he found the remains of an employee selection test-paper, for a research assistant. Part of it is shown below, although the bit marked "TOP SECRET- New 3D circuit board for new 3D dvd player" was a bit difficult to read properly, so I haven't included it in the scam.

Anyway, this part of the test, asks the candidate to work out the resistance between point A and point B which are at the opposite corners of an open cube, where each edge of the cube is made from a 1 ohm resistor.

Apparently this helps to improve the 3D effect of movie films when played on a new dvd player to be launched later this year...........something to do with circuit topology being important....

How would you fare in this part of the test.

Cheers

Don

5/6 of an Ohm.

You can cheat and use a Spice simulator, but imagine a current of an amp flowing into A and out of B and think about the voltage at each corner of the cube.

Ohms law is V=IR, volts is equivalent to current times resistance.

Each resistor connected to A must carry 1/3A, each arriving at B similarly 1/3A. The 6 inbetween must be carrying 1/6A. Following any path from A to B we get 1/3A-1/6A-1/3A in 1 Ohm resistors, so the voltages go 1/3, 1/6, 1/3 for a total voltage drop of 5/6.

Paul

My resistance is low

5/6 ohm is right. Well done Paul.

I had visualised it as follows....

At A, the circuit splits into 3 x 1 ohm resistors which are thus in parallel; each of which then splits into 2 x 1 ohm resistors (which thus gives 6 in parallel); before the pairs re-join to give a further 3 x 1 ohm resistors (again in parallel).

Hence the total resistance is 1/3 + 1/6 + 1/3 = 5/6 ohm

Anyway Paul, I think you are halfway to getting that job, but I do have one or two more snippits from the dustbin......

Cheers

Don

A dicey question

A dice has the numbers 0; 1; 2; 3; 4; and 5 on its faces.

It is thown until the succesive numbers shown at each throw, add up to more than 12

eg 2 + 5 + 0 + 4 + 4 > 12 = 15

What is the most probable score in such a game?

Cheers

Don

Forgive me, I cannot read

Alex, could I introduce you to "Janet and John" ?

Or help in some other way?

Cheers

Don

Don,

I'll stick my neck out and go for 13. I await the right answer.

Interestingly, I get 13 approx. 1/3 of the time, 14 approx. 1/4 of the time, and 15 approx. 1/5 of the time.

cheers,

Dan

[This message was edited by Dan M on WEDNESDAY 21 January 2004 at 22:12.]

Naim Technician Exams-Part 2

Since Paul R was the only candidate to attempt Part 1, he is currently a clear runner for the job.....

Part 2 involves calculating the precise length of thin, pure gold wire, needed to make an inductor.

The diagram below should help. It shows a spiral inductor, wrapped as a 10-turn helix around a cylindrical tube, 4 inches diameter and 9 inches long. The ends of the wire coincide with a common element of the cylinder ie they start and finish at the ends of the same longitudinal line.

So......what length of this precious wire is needed?

Cheers

Don

I'll stick my neck out and go for 13. I await the right answer.

Apologies Dan, I should have put you out of your misery a long time ago (in the nicest possible way.....)

You are spot-on with 13

Cheers

Don

10 * sqrt( pi*pi*16 + 81/100 )

-Dan

Dan,

That's not the answer I have.

So, either one of us, or both of us are wrong

Cheers

Don

Dan has come up with the answer I would have given. Obviously break the problem up in to 10. We have 1 coil round a 0.9 inch length tube, with diameter 4 inch.

If you unravel the tube then you would have a right angled triangle, with the 2 shorter sides being 0.9 inches long and pi*4 inches (circumfrence of a circle = pi * diameter). Using pythagarous's (sp?) theorum you get sqrt( pi*pi*16 + 81/100 ), and then multiply this by 10.

The only thing that I could see wrong with this is that if you unravel the tube you don't get a right angled triangle, but then you wouldn't have uniform coils.

Don,

I agree with Dan's answer too.
((4 x pi x 10)^2 + 9^2)^0.5 inches

Matthew