Brain Teaser No 1

Who'd be a dealer ?

You are a second hand hifi dealer. Your first customer of the day buys a CD player for $250 and pays with a cheque, changes his mind and asks for a $150 amp instead. With an empty till you cash the cheque with the butcher next door and give the buyer a $100 bill as change.

The cheque bounces and you have to borrow $250 to pay the butcher. The amp cost you $110. How much money have you lost.

Cheers

Don

What gear have you got....


Here are four cog wheels. The largest cog has 86 teeth, the next cog has 25 teeth, the third cog has 15 teeth and the smallest cog has 12 teeth. (the diagram is only schematic !)

How many revolutions will the largest cog make before all the cogs return to their original positions ?


Cheers

Don

Dear Don,

I must say that I feel some what insulted that you would assume that I had such foolish business practices but I would find myself $210 out of pocket.

I counted that the large cog went round 150 times before they had all returned to the starting position.

cheers

Matthew

Lynn, Matthew,

Yep, $210 and 150revs

Cheers

Don

A Monumental Problem.

A 10 tonne statue with a flat bottom, has to be placed upon a pedestal by crane. There is no way of lifting the statue except by slings under the base.

How could you arrange to set the statue on the pedestal and get the slings out ?

Cheers

Don

OK

Whilst the concept of the monumental statue gradually sinks in, here is a 'tide me over' teaser.

I have 50 stadard US coins which total one dollar.

Which coins do I have and how many of each?

Cheers

Don

Don,

There is more then one solution to the coin problem, do you want both?

Matthew

There is more then one solution to the coin problem, do you want both?

Matthew, yes please.

Suggest you post the 3-denomination solution first, then after a reflective pause, the 4-denomination solution.

BTW, you were meant to only find one solution and post that with glee !!! I was then meant to respond that most mathematicians/puzzle solvers were lazy and assume there is only one unique solution. Archaeologists fall into a similar group....



Cheers

Don

[This message was edited by Don Atkinson on TUESDAY 15 April 2003 at 21:34.]

Don,

The 3 denomination solution is...

40 pennies
8 nickels
10 dimes

Matthew

Errrr yes, 2 dimes

quote:

---

Originally posted by I was dobbin once!:
The other solution is:
1x45c/2x10c/2x5c and a 25c piece (apologies for my unfamiliarity with the colloquial names of US coinage!)

---

Aha! The famous 45c coin, colloquially named the What-are-you-on.

ice ice baby

Cool answer man, cool answer.

Cheers

Don

The magic of spellcheckers is lost on mathematicians....

the last few posts have claimed:-

40+8+10=50 (Matthew)
1+2+2+1=50 (I was dobbin once!)
and introduced a new 45 cent coin to the USA

I see this thread heading rapidly downhill.....which reminds me of another teaser

Tommorow

Cheers

Don

Hello all,

I've been traveling for a few weeks and see there's been lots of fun teasers I have missed. Here's a quick teaser -

what's the square root of 12345678987654321?

Hint 1*1 = 1, 11*11 = 121.

-Dan

Dan,

That hint was too big!

Matthew

quote:

---

Originally posted by Matthew T:
Dan,
That hint was too big!
Matthew

---

...especially with this nifty little onscreen calculator.

quote:

---

Originally posted by Matthew T:
That hint was too big!

---

well, in my defense, it was supposed to be a quickie

Dan

Dan,

I think you'd best publish the answer for the benefit of all the browsers who don't have excell or large calculators but do enjoy nine pin bowling....

Cheers

Don

Don,

Very well, the answer is of course 111,111,111. Now how about that "downhill" teaser?

cheers,

Dan

The downhill teaser.

The usual idealised conditions apply. Perfect sphere; smooth slope; no friction losses but no sliding/slipping; zero drag; no wind resistance etc etc.

I have a ramp 64m slope length. The sphere was released at the top of the slope with zero speed and is now rolling down the slope and passing the 16m mark.

In terms of journey time, how far down the slope is the sphere ?

Usual rule for simple ones....must be dome in the head....

Cheers

Don

quote:

---

In terms of journey time

---

I believe the sphere is half way. I could be wrong though since I resisted the urge to reach for a pencil.

cheers,

Dan

ok gents, one from my daughter's geometry class.

What time does Ms Lynn have to hand in her geometry homework ?

One is doing one's best to help one's friend with one particular problem.

Now one could provide one's rather ugly trig solution based on the use of one's calculator or one's excell spreadsheet.

Or one could provide one's slightly more elegant looking solution based on trig functions and geometry and one's rather hazey recollection of how to get these things to work together harmoneously.

Or one could wait just a little while to see whether someone else has a better idea.

Cheers

Don

Don,

The rolling sphere will be half way down.

Lynn,

Here is my solution.

The cecnter of the circle will sit on the cross point of the lines which cut each corner of the triangle in half (equal angle between the two sides of the triangle).

If we take the 3 cm side we can see the the angles of these line between the 5 and 4 cm side are, respectively

arctan (4/3)/2 ( which can also be expressed as arcsin(4/5) and arccos(3/5) and 45 degrees (right angle triangle thankfully)

We can then find the intercept of these lines by solving the following equation

3 = x1 + x2

where

x1 = r

and

x2 = r / (tan (arctan(4/3)/2))

tan (a/2) can be represented as sin a/(1 + cos a) and therefore this gives

3 = r + r /(sin(arcsin((4/5))/(1 + cos(arccos(3/5))))

3 = r + r / (1/2)

or

r = 1cm

Matthew

Non-trig solution
I have a geometry solution that is difficult to describe without pencil and paper, but here goes:

Given right angled triangle (a,b,c) = (3,4,5), the center of the circle lies at (r,r). The tangent points for sides a and b lie at (0,r) and (r,0). With me so far? The remaining lengths along sides a and b are (a-r) and (b-r). If you sketch this out you'll see that c must equal (a-r) + (b-r).

So a + b - c = 2r

r = (3+4-5)/2 = one

cheers,

Dan

p.s. so Don is 1/2 way the correct answer?

a little help from my friend.....

Dan,

Perhaps this diagram would help your solution along. The critical bit is using similar triangles to show Bx = Bz =(a-r) and likewise Ay = Az = (b-r)

Then its easier to justify c = (a - r) + (b - r) etc

Its also worth re-iterating what Mattew said about the centre of the circle O being defined by the bisectors of the angles A, B and C.

I have also drawn in the third radius, oz because this helps identify the similar triangles etc

Hope all this helps Ms Lynn with her homework.....

Cheers

Don