Brain Teaser No 3

Here's an educational one for er, everyone:

It's WWI, and it's a big battle, 18 hours per day alternated with 6 hours of rest.
You calculate the depletion rate of your soldiers and it's about 36 square meter per hour.
What should be the inflow of new soldiers, at a constant rate 24 hours per day,
to keep your usable soldiers at a constant density.

Yes, something with increasing the sides of the polygons. There is another one I vaguely recall, about showing that the square root of 2 is not rational. If anyone can reconstruct it...

hint: assume that sqrt(2) is a fraction and show this leads to contradictions.

Someone showed me this one already so I won't spill the beans.

I am in awe of the extent of a theoretical 12 year old's mathematical savvy.

On a point of information, I never run up escalators or anywhere else really...

quote:

ok, here's a little geometry one so Don can join in.

Not at 11 o'clock at night he can't..........

But yes, I can see that Archimedes could draw a hexagon inside the circle made up of six equalateral triangles each of side 1 and thus deduce that the circumference of the circle was a little bit more than three times the diameter, the diameter being 2, ie the co-linear sides of two of the triangles.

He could also deduce that the circumference was a little bit less than 6/rt3 given the side of the external triangles (L) could be deduced from Pythagoras' theorem based on L^2 = (L/2)^2 + 1
giving L = 2/(sq rt 3).

Well done Archimedes......and Alex

Cheers

Don

quote:

Someone showed me this one already so I won't spill the beans.

Hmmmmmm,

my initial thoughts are something like...

If sq rt 2 IS rational it can be expressed as a fraction such that (sq rt 2) = p/q where p and q are integers and primes (ie fully reduced)

square both sides

2 = (p^2)/(q^2)

2(q^2) = (p^2)

But this would require p^2 to have a factor of 2, which it can't because p is a prime and p^2 is simply that prime muliplied by itself.

So there is a contradiction

Which means that the intial assumption (ie that sq rt of 2 is rational) must be invalid

But I'm not totally convinced this is rigorous enough........I need to give it more thought.

Cheers

Don

quote:

I am in awe of the extent of a theoretical 12 year old's mathematical savvy.

Awe come on Nigel....

Let N = number of bricks in wall

N/9 = No of bricks Nigel lays per hour
N/10 = No of bricks Rupert lays per hour
N/9 + N/10 - 10 = No of bricks per hour they lay together

N/(N/9 + N/10 - 10) = 5 (five hours = time to build wall)

Solving for N is simple even for a 12 year old (it isn't even a quadratic)

Oh, go on then.....

90N/(10N + 9N - 900) = 5
90N = 5(10N + 9N - 900)
90N = 50N + 45N - 5x900
N = 900

Cheers

Don

quote:

Here's an educational one for er, everyone:

Soldiers are wasted at a steady rate of 36 sqm per hour for 18 hours then none are wasted for a period of 6 hours.

Reinforcements arrive at a steady rate over a full 24 hours.

On average you would want them to arrive at 27 sqm per hour.

But that would result in a steady decline during the 18 hour fighting period and a surplus during the lull.

Of course, if the reinforcements arrived at the same rate as the fighting attrition, ie 36 sqm per hour, but during the lull they were "returned to the rear", then the numbers on the battlefield would remain more or less constant.

But somehow, I don't think this is what you were after........

Cheers

Don

quote:

But I'm not totally convinced this is rigorous enough

If you show that squares of odd numbers are never even and that p isn't 2 then I think you're done.

Paul

Aahh...

quote:

If sq rt 2 IS rational it can be expressed as a fraction such that (sq rt 2) = p/q where p and q are integers and primes (ie fully reduced)

square both sides

2 = (p^2)/(q^2)

2(q^2) = (p^2)

But this would require p^2 to have a factor of 2, which it can't because p is a prime and p^2 is simply that prime muliplied by itself.

So, continuing....

p^2 has to be even, (since it is 2x something ......something = q^2)
This can only be true if p itself is even (* see below)
If p is even, then p^2 is divisible by 4
In which case q^2 (and therefore q) must also be even

This means that both p and q are even. This is a contradiction of the intial assumption
Hence sq rt 2 is not rational

(*) the product of any two numbers is the product of the factors of both numbers.
If neither number has a factor of 2 (ie they are both odd) then the product must also be odd, since it won't have a factor of 2
Hence the square of an odd number must also be odd.

I feel more satisfied with the above additions........

.....but it doesn't look elegant.....

Cheers

Don

Train sets

Nigel has just qualified as a train driver.

His first job is to use his red loco to swap the positions of blue coaches A and B and return the loco to its original position.

The loco can pass through the tunnel but the coaches can't. Each end of the loco can be used to push or pull coaches and the two coaches can be coupled together if required.

Obviously Nigel wants to complete the task efficiently.

How can he do this in the smallest number of loco "moves"

A "move" is any movement of the loco between stops.

Stops includes : stopping to change direction; stopping to couple or uncouple the loco or coaches to each other. Moving the points/switches is not counted as a "move".

Flying switches are not permitted, ie ones where the loco pushes or pulls a coach over the points which move swiftly to send the loco one way and the coach another way.

Anybody care to help Nigel ?

Cheers

Don

Not a Rubik Cube

Just to provide a little light entertainment while you get the train-sets out....

What is the smallest number that is the sum of two cubes in two different ways. Use positive integers only.

Cheers

Don

I'm back! for a short moment at least. I wouldn't look at prime numbers there, you'd end up with top and bottom of the fraction being products of primes anyway. The key thing is to simplify the fraction till either top or bottom is even but we don't know which. So if we again have the situation 2(q^2) = (p^2), we can deduce that p is even and q is odd. So we replace our fraction by 2*P/q, (we don't know if P is odd) leading to 2(q^2) = 4(P^2). Now q is even, which can't be right.

quote:

What is the smallest number that is the sum of two cubes in two different ways. Use positive integers only.

10^3 + 9^3 = 12^3 + 1^3 = 1729

It's worth googling 1729.

Paul

quote:

It's worth googling 1729.

Never a dull moment.........

It never occured to me to Google 1729. I had forgotten many of the stories surrounding this number. Some interesting articles. Thanks Paul

Cheers

Don

The Great Dash!!

An easy one, just to get the old brains in gear for the weekend.

Bill and Ben race each other over 100m. Ben wins by 10m. They race again, but to even things up, Ben starts 10m behind the start line, while Bill starts on the start line.

Assuming they both run at the same constant speed as in the first race, who wins?

Cheers

Don

gie's a break - I'm still on and off wrestling with the bloody train one !

I haven't even found bad solution never mind a good one

I found a solution for the train switching thing in 18 moves. It's easy to miscount though.

The general idea was to first get behind(on the right side of) the two coaches near B. Then you are free to maneuver the two coaches. You make sure the coach that has to go to A is on the straight part bottom left. Then you pull the other coach to B and continue through the tunnel, pick up the coach bottom left, push it to its place, and go home.

quote:

Originally posted by Don Atkinson:
The Great Dash!!

An easy one, just to get the old brains in gear for the weekend.

Bill and Ben race each other over 100m. Ben wins by 10m. They race again, but to even things up, Ben starts 10m behind the start line, while Bill starts on the start line.

Assuming they both run at the same constant speed as in the first race, who wins?

Cheers

Don

Ben wins as he will be able to run just over 111m in the time taken for Bill to complete the 100m.

quote:

I found a solution for the train switching thing in 18 moves. It's easy to miscount though.

I agree that "counting" what constitues a "move" is difficult, so whether its 16, 17 or 18 moves could be debatable - when I first saw this puzzle, the answer quoted was 16....

Hovever, your general description of how to do it is spot on - terrific.

Cheers

Don

quote:

just over 111m

ltd, well done - and nice to see a new face having a go.

Are we all agreed that "just a little over 111m" is equal to 1/9m ??

Cheers

Don

Alex,

Do we need a bit of encouragement on the battlefield?

Cheers

Don

That Little Train Set
Here are my 16 moves and a labled diagram to help.
Note The Blux Coach is labled X and the Yellow Coach is labled Y

1 Loco to A and hook up X
2 Loco + X to B and stop
3 Loco + X to S2 and unhook X
4 Loco to B and stop
5 Loco circle clockwise to B and hook up Y
6 Loco + Y to S2 and hook up X
7 Loco + X + Y to B and stop
8 Loco + X + Y to A and unhook X
9 Loco + Y to B and stop
10 Loco + Y to S2 and unhook Y
11 Loco to A via tunnel and hook up X
12 Loco + X anticlockwise to B and unhook X
13 Loco to S2 and hook up Y
14 Loco + Y to B and stop
15 Loco + Y to A and unhook Y
16 Loco to S1

Cheers

Don

Admin BUMP