Brain Teaser No 1

the number 5000 can now be represented in two ways, but that shouldn't be a problem.

Now all I need to do is find a REAL dealer where 552s come in at £10k and cdsiis come in at £5k so that I can put your answer to the test!

Cheers

Don

I have three cider jugs, one holds 8 gallons, one 5gallons and the third holds 3 gallons. The large jug contains 8 gallons of cider, the other two are empty. I have no other jugs, bowls, bottles etc and the jugs I do have, have no markings other than the 8, 5, 3 gallon marks respectively.

I need to share the cider with my friend so that we each have 4 gallons.

How should I set about dividing my 8 gallons into two equal quantities?

Cheers

Don

The first three columns represent the contents of the 8, 5, and 3 gallon drums, respectively. The fourth column shows the operation to be performed (e.g. 8 -> 5 means pour contents of 8 into 5 until either 8 is empty or 5 is full).

8 0 0 : 8 -> 5
3 5 0 : 5 -> 3
3 2 3 : 3 -> 8
6 2 0 : 5 -> 3
6 0 2 : 8 -> 5
1 5 2 : 5 -> 3
1 4 3 : 3 -> 8
4 4 0

(Hic!)

JD

JeremyD

A neat presentation of the solution, IMHO.

However, only after your last line (440) had changed into (000) would I expect to see a 'hic!'

Cheers

Don

read on.....

Do the following sums as quickly as you can:-

1 - 1 = ?

4 - 1 = ?

8 - 7 = ?

15 - 12 = ?

Pretty sure you got them all right ?

OK. Now for the next bit. Without 'messing about' or 'thinking too hard' pick any number between 12 and 5.

Now you are probably wondering what this is all about. You're not the only one....but be honest...

Chances are, the number you picked is 7.

Any idea how I knew ??

Cheers

Don

[This message was edited by Don Atkinson on SATURDAY 20 July 2002 at 09:19.]

Since I'm sure the answer has nothing to do with the fact that the sum of the answers is 7, my answer is, "No."

BTW, I chose 6 - it's the lazy choice.

A maths teacher once asked my class of about 30 to pick a number between 0 and 9. I think more than a third of the class chose 7. Most odd...

JD

JerremyD and Omer and any others who might try it.

Most people choose 6,7 or 8, with the highest proportion choosing 7. Of course not everone does. My eldest (25 years) chose 2....but then you always get one like that......

Why choose 6, 7 or 8 ?

Well, the best explanation i've heard is that you've just done four subtractions in your head and the old brain has gone into 'instinct' mode. It then sees 12 and 5, where the difference is 7, and despite knowing that it no longer has to do a subtraction, it does so, instinctively.

Of course, some people 'see' the difference only as an approximation, and go for 6 or 8. Only a few go for the 'average' ie 8, 9 or 10. Same applies to the top end 11 and 12.

Of course, only if 100+ members posted there 'choice' would we know if there was any truth in this.

Cheers

Don

Don: the old brain has gone into 'instinct' mode. It then sees 12 and 5, where the difference is 7, and despite knowing that it no longer has to do a subtraction, it does so, instinctively.

I'm not sure mine does, although I'd be very pleased if it did: I'm an arithmetically challenged ex-ish maths student. I first began to like maths when letters started replacing numbers.

It's only through watching Countdown that I've become respectably competent at mental arithmetic.

JD

I'm an arithmetically challenged ex-ish maths student

I think most of us are, if the truth be known. (arithmetically challenged, that is)But it does get irritating if the arithmetic keeps buggering up the algebra, trig, statistics, theory of games, or whatever. So eventually, most of us master the arithmetic bit. Bit like English grammar......possibly.

Oh, BTW, i'm sure that the modal choice for arithmetically challenged professors of mathematics is 6. No proof, just a hunch !

Cheers

Don

Mick was standing at Swindon station waiting for his friend, when two express trains of the same length roared past in opposite directions, each at its own constant speed. With his latest acquisition to that famous watch collection, Mick timed them and noted the faster one took 6 seconds to pass him and the slower one took 8 seconds.

Just then, Mick saw his chum Marcus, who had come down from Wales for a spot of shooting, 30 yards along the platform. (that’s where Mick saw Marcus standing, not where they were going for a spot of shooting !). Mick told Marcus that he had been standing exactly in line with the point at which the front of the two trains coincided.

"What a coincidence", cried Marcus, "I was standing exactly in line with the point at which the ends of the trains coincided!".

Mick was over the moon. He was now able to work out the length of the trains.

Can you ?

Cheers

Don

I'm not sure if this is the same logic as James used, but here goes:-

Let the time taken between the front of the trains passing, and the back of the trains passing, be T. Length of train = X.

The faster train has travelled X plus 30 yds, at say speed S in time T.

Therefore T = (X+30)/S

The slower train has travelled X minus 30 yds, at speed 0.75S in time T.

Therefore T = (X-30)/0.75S

Solving this gives X = 210 yds.

STEVED

First and foremost, well done ! The length of each train is indeed 210 yards.

How James held his mind together doing that arithmetic and in such a short time, amazed me.

HOWEVER, (you can now sense the smug bas*ard creeping out, can't you!)

However, you could try :-

The ratio of the speeds is 4:3

Let the length of each train be L yards

In travelling from 'nose to nose' to 'tail to tail' the faster train will travel L + 30 yards and the slower train will travel L - 30 yards

Since distance travelled is proportional to speed we can write

(L + 30)/(L - 30) = 4/3

which develops into

3L + 90 = 4L - 120

and L = 210

Of course, this is VERY similar to steved's version, so the smuggness wasn't justified at all.

Who said "bring back the 11+" ?

Cheers

Don

Does this mean it’s my turn to pose a question?

It certainly does, but you would be welcome to post questions anytime, the more the merrier etc etc

I was talking to a guy down the pub the other night and he set me a couple of teasers that I hadn't heard before. Trouble is, he didn't know the answers, and I have sorted them out either yet....could take a little while!!

Cheers

Don

quote:

---

Trouble is, he didn't know the answers, and I have sorted them out either yet....could take a little while!!

---

Give us a go anyway!

Paul

England, France and Germany played each other in a football tournament.

England beat France, France beat Germany and Germany beat England. (Heard it before have you !)

England scored the most goals
France had the best goal average (for / against)
Germany had the best goal difference (for - against)

Fewer than 40 goals were scored in the tournament.

What were the scores in each game.

Perhaps we had best get the 'world cup' contributors over onto this one ?

Cheers

Don

England 6 France 5
France 2 Germany 0
Germany 15 England 11

Goals for:
England 17
France 7
Germany 15

Goal Difference
England -3
France 1
Germany 2

Goal Average
England 0.85
France 1.167
Germany 1.1538

Total Goals 39

Not sure if this is unique, but given that the goal total is just less than 40 I suspect that it may be.
Had to do it by informed trial and error. Is there a mathematics for solving simultaneous inequations?

Just a quick one. Being a luddite, I still get my full cream milk delivered in a traditional glass bottle. If I were to measure the hydrostatic pressure in the bottom of the bottle, with the cream separated, and then to vigorously shake the bottle (without spilling any) so as the cream is well dispersed, would the hydrostatic pressure have changed? If so would it have increased or decreased?

Simon

There are numerous factors which could effect the outcome.

The effect of going from a creamy mensicus to a full fat milk mensicus could change the pressure, not sure which way though.

The volume of milk and cream could be different to creamy milk

Anyway, as far as I can remember once shaken the top tends to pop up a little which would suggest higher hydrostatic pressure.

If you shake it really hard it would also warm up the milk and therefore increase the pressure until the top taken off.

Just some thoughts.

Matthew

Not sure if this is unique, but given that the goal total is just less than 40 I suspect that it may be.
Had to do it by informed trial and error. Is there a mathematics for solving simultaneous inequations?


As I said, before setting the question, the guy down the pub didn't know the answer and I hadn't sorted it out either. However, your answer certainly fits the bill. Whether its unique.........As for a formula...........

Sorry for the delayed reply.....long holiday and busy, busy , busy back at work.

Cheers

Don

OK, a variation on 'The Tower of Hanoi'

Using coins instead of rings. And English coins at that, which makes it even harder for Omer, Vuk and everybody in Euroland !!

Start with a pile (tower) of coins comprising at least one each of the 1p, 2p, 5p, 10p, 20p, 50p, and £1 coins. No coin is allowed on top of another that is smaller in diameter. (for our non-English friends this means the 5p coins are on top, then the 1p, 20p, £1, 2p and 50p)

The object of the game, as with the original puzzle, is to move the tower of coins to a new position, one coin at a time. At any stage, there can be up to three piles (or towers). ie the original pile, the final pile and an intermediate pile. In no pile, at any time, can any coin be above another of smaller diameter.

I did this with a pile of coins yesterday and found that the minimum number of moves equalled the value of the pile in pence. Today I doubled the number of coins by adding some 1p, 5p, and 50p coins totalling less than £3 and repeated the game. Again the minimum number of moves equalled the value of the new pile in pence.

How many coins were in the pile in the first game ?

Cheers

Don

Don,

Welcome back. Good to see this thread back in action.

My stab at this is 12 coins. Will go through the working once others have had a guess.

Matthew

Mathew,

Thanks, its good to have a bit of time again.

My stab at this is 12 coins. No comment, yet......

Will go through the working once others have had a guess.All guesses are welcome.

Better still, why not start by trying to move a few coins (not you Mathew....)to see if there is a simple formula for the minimum number of moves.......but if this fails, have a guess anyway!

Cheers

Don

Listening to 'the War of the Worlds' last night (Arkiv/LP12/Ekos/lingo/linto/52/supercap/135s/System5 all sitting on Base except the LP12 which sits on Audiotech and beats Mana to hell, or Mars, and back - I have never heard Mana but I jusy KNOW Audiotech and Base are streets ahead) I got into my mind that catchy phrase "the chances of anything coming from Mars are a million to one, he said"

So, whilst falling asleep, the old mind started drifting. The figure 1,000,000 floated by and I began to realise that there is only one pair of numbers, neither of which contain any zeros, but which, when multiplied together make a million.

So, as a little light relief from relocating the Tower of Hanoi, what are these two numbers?
Cheers

Don

PS (ie Post Script, not Paul Stephenson !) Don't loose too much sleep over the Base & Audiotech v Mana, I don't. If the Fraim had been invented before I bought the Base, I would have have bought the Fraim)

Well I cheated with an exhaustive search on the sound basis that it wouldn't require more than 1000 tests...

So from the 24 pairs of numbers that multiplied together make 1000000 I pick 64 and 15625.

Curiously a normal PAL TV scan line lasts 64 micro seconds, so the horizontal frequency is 15625Hz.

Paul

The minimal solution to a conventional ToH problem takes 2^N-1 moves, where N is the number of discs.

In our case we have 7 discs which will take 127 moves yet have a value of 188.

If we add another disc of the same size as an existing one then it only adds a number of moves equivalent to the number of times the original disk moves. So if we add another instance of the largest then we just require one more move, of the smallest (of 7) then 64 more moves are required.

The order of the coins is (smallest first) 5p, 1p, 20p, 100p, 10p, 2p, 50p.

So we're looking for solutions to,

127+64a+32b+16c+8d+4e+2f+g=188+5a+1b+20c+100d+10e+2f+50g

and the total coins is 7+a+b+c+d+e+f+g, any of a-g can be 0.

There are rather a lot of possible solutions. The problem requires us to find a pair where the number of coins in one is twice that of the other, the only coins that have been added are 1, 5, or 50 and the total added value is less than 300.

Thanks to the magic of computers I think there were 12 coins in the first pile and 24 in the second composed as follows,

coin
   1 :   1   7
   2 :   1   1
   5 :   3   4
  10 :   1   1
  20 :   3   3
  50 :   2   7
 100 :   1   1
=================
count:  12  24
value: 288 549

No warranties etc... I cannot prove my program is correct and now over to the mathematicians to demonstrate a simple analytical way of proceeding!

Paul

Well I cheated...So from the 24 pairs of numbers that multiplied together make 1000000 I pick 64 and 15625.

Getting the answer is all that matters, well done.

Of course........... there is a neater way (IMHO) of getting the answer........ and explaining it (IMHO).......... and showing that it is the ONLY pair of numbers that fit the criterea........... (nothing humble about this last bit)...........

Cheers

Don