A Fistful of Brain Teasers

Aaron 20,2,3,25,20,1

Bertie 2,50,10,5,3,1

Chuck 20,10,10,1,25,5

any prizes for guessing the order?!

Innocent Bystander posted:

Aaron 20,2,3,25,20,1

Bertie 2,50,10,5,3,1

Chuck 20,10,10,1,25,5

any prizes for guessing the order?!

You can give it a go - for the Kudos.....

The Baron's Treasure

The Anglo-Austrian Baron Von Dane-Fraydenegg kept his gold treasure (fairly got by moderating various fora) in a treasure house. In each room of the treasure house were as many chests as there were rooms. In each chest there were as many gold coins, as there were chests in that room. All the gold coins were of the same size and value.

When he died, the Baron's will was that his surgeon-barber should receive one chest of gold coins. The remaining coins were to be divided equally between the Baron's three sons .

The three sons were proud and fierce men who would definitely resort to bloodshed if the coins could not be divided equally.

The question is simple: -

a) Was there blood shed.
b) Was there no blood shed.
c) Is there no way of knowing whether there was blood shed or not.

The proof is the real requirement, not simply a one in three guess.

Answer = c) : it depends on the number of rooms.

X rooms
X^2 chests
X^3 coins

Surgeon 1 chest = x^2 coins
Brothers (x^3-x^2)/3 = whole number

2 rooms = no 1.33 coins each) 3 rooms = yes (6 coins each)

There wil be no bloodshed (unless there was only 1 room and the sons turned on the surgeon barber).

r = no of rooms and no of coins per chest

r^2 = total no of chests

r^3 = total no of gold coins

barber receives 1 chest = r coins

balance of coins = r^3 - r = r(r^2 -1)

Obviously if the no. of rooms is a multiple of 3 then r(r^2 -1) is divisible by 3 otherwise, if r is not divisible by 3 then (r^2 -1) is divisible by 3.

Either way the product of r(r^2 - 1) must be divisible by 3 so that the coins can be divided equally between the 3 sons.

Trust me to rush it and get it wrong. Indeed the surgeon gets x not x^2 coins, and Sjbabby is right

!!!

But it's possibly the way many of us went first time around, so you're probably not alone ! 

sjbabbey posted:

There wil be no bloodshed (unless there was only 1 room and the sons turned on the surgeon barber).

r = no of rooms and no of coins per chest

r^2 = total no of chests

r^3 = total no of gold coins

barber receives 1 chest = r coins

balance of coins = r^3 - r = r(r^2 -1)

Obviously if the no. of rooms is a multiple of 3 then r(r^2 -1) is divisible by 3 otherwise, if r is not divisible by 3 then (r^2 -1) is divisible by 3.

Either way the product of r(r^2 - 1) must be divisible by 3 so that the coins can be divided equally between the 3 sons.

 

Neat solution sjb.

I normally expand the expression (r² - 1) to give (r-1)(r+1)

which together with "r" gives the number of coins as the product of r(r-1)(r+1)

which re-arranged as (r-1)r(r+1) is more obviously the product of three consecutive numbers

one of which has got to be divisible by 3

Yes I'd forgotten that simplified proof.

There is a less elegant proof that r^2 -1 is divisible by 3 when r is not itself divisible by 3.

Let r = 3n - 1

then r^2 =  (3n-1)*(3n-1)     =     9n^2 - 6n + 1  

So r^2 -1 =     9n^2 - 6n       =    3n(3n -2)

Let r = 3n - 2

then r^2  = (3n - 2)*(3n - 2)  =  9n^2 - 12n + 4

So r^2 -1  = 9n^2 - 12n + 3   = 3(3n^2 - 4n + 1)

The Lord of the Wings Bookworm

I was re-arranging my bookshelf the other day. My three volumes of “Lord of the Wings” needed dusting but I noticed they were in the “correct sequence” as you might expect and as shown in the diagram (although in practice they were tightly packed together - you could almost say "stuck together)

However, much to my dismay, I discovered that a bookworm had tunnelled its way from Page 1 of the trilogy, right through to the last page of the trilogy.

Now the three volumes are identical in size and construction. They are hard-backed, with the front and back each 1/8 of an inch thick. The pages are together three inches thick in each volume.

How long was the tunnel that had been bored by the bookworm ?

I make that 1 1/2 inches

3.5"

notnaim man posted:

I make that 1 1/2 inches

I'm going to give you 9/10

I can see that you recognise the solution, ie where the first and last pages are.

You probably got slightly confused with the thickness of each volume being three inches, not each volume being one inch.

"The pages are together, three inches thick in each volume"

Innocent Bystander posted:

3.5"

10/10

Don Atkinson posted:

Innocent Bystander posted:

3.5"

10/10

Of course, anything longer could be correct, as there was no indication that the worm burrowed perpendicular to the covers or even in a straight line, so actually the only correct answer is at least 3.5”!

A rope is passed over a pulley which is attached to the rigid roof.

At one end of the rope is a weight.

The weight exactly balances a man hanging onto the other end of the rope.

ie it’s in static equilibrium.

The man starts to climb the rope. Don't rely on the picture, most people climb a rope using their feet together.

What happens ?

The rope is fully flexible and the pulley is frictionless. I therefore don’t think it matters whether the rope/pulley contact is frictionless or not. But if you consider this aspect to be important, say so and allow for it in your explanation.

The weight will begin to rise to maintain the equilibrium of the system i.e. to ensure that the moments (force x distance) on either side of the pulley remain equal.

O f course, neither of the above seems IMHO to take acceleration into account.

But I suppose, that if the man climbs slowly, then it might be possible to ignore acceleration.

But when he stops climbing, no matter how slowly he was climbing, I have this inner feeling that the system as a whole continues to move, unless a little bit of friction exists between the rope/pulley or in the pulley bearing !

I have £1000 in £1 coins.

I also have 10 bags in which to put those coins.

Each bag will be sealed and labelled showing the number of coins therein.

I wish to distribute the coins to those bags, such that going forward I am able to simply pick up one or more bags and without opening them or redistributing the coins further, be able to take any sum of £££ from £1 to £1000.

How should I distribute the coins.

I only need to be able to make one such transaction, not multiple transactions.

Thats the classic 225 stages used, for example, balance weights, just with an initial single unit:

1,2,2,5,20,20,50,200,200,500

Hi IB,

I can see where you are coming from with the "weights" analogy, but.....

.....which bag, or combination of bags, from your list would you hand to me if I asked for (say) £11 ?

or perhaps (say) £67 ?

The idea is that I know how much is in each bag (I put the coins into the bags and labelled them accordingly) and in future need to be able to simply pick up one or more bags eg in a bit of a hurry !......so as to take ANY sum of whole pounds from £1 to £1000

Cheers, Don

Having not used physical weights with balances I misremembered the sequence, and should have stopped to check but didn’t as that was 10 bags. Correct sequence tmake any number is of course 1,2,2,5,10,10,20,50,100,100,200,500

However that would require 2 bags too many, so I’m looking for the hidden twist, but so far to no avail, other than wondering about your need to state that you only need to make one transaction, which I had taken as read. I suppose you could have 3 single coins in your pocket, so not needing the first 2 bags, but I think that is outwith the meaning.

No, there is no hidden meaning. It’s just straightforward. 

10 bags contain all 1000 coins. We just need to put the right number of coins into each bag so that by selecting one or more bags you can pick up any amount of £££ from £1 to £1000.

If the bags were then replaced, you could select some other amount by selecting a different combination of bags ........etc etc

so, no hidden meanings, no twist of words, just a straightforward puzzle !

1, 2, 4, 8, 16, 32, 64, 128, 256 and 489