Brain Teaser No 1

We already know that the maximum number of footballs, cricket balls, golf balls or marbles that we can stack so that each one just touches each of the others, is five, and that applies to any combination of footballs, cricket balls, marbles etc as well. And Matthew T has clearly got a spread sheet for working out the size of the fifth one, given the size of the other four.


Oh, it also applies if one or more of the balls surounds the others, like a bubble, or a bowl.

So, the magic formula ?

'curvature' is the inverse of 'radius' so a sphere of radius R has a curvature of 1/R.

let the curvatures of 5 spheres be a, b, c, d, and e respectively. If a sphere encloses the others, its curvature is negative, eg -b.

3(a^2 + b^2 + c^2 + d^2 + e^2) = (a + b+ c + d + e)^2

Given 4 radii, the 5th can be found by substitution and solving the ensuing quadratic.

Trivial. But I am impressed with Matthew T's spreadsheet!!

Cheers

Don

There came three Dutchmen of my acquaintance to see me. Being lately married they brought their wives with them. The men's names were Hendrick, Claas and Cornelius. The women's Geertrick, Catriin and Anna, but I forget the name of each man's wife.

They told me they had been to market to buy hogs. Each person had bought as many hogs as they gave shillings for each hog. hendrick bought 23 hogs more than Catriin and Claas bought 11 more than Geertrick. Likewise, each man laid out 3 guineas more than his wife. I desire to know the name of each man's wife.

Note, a guinea was 21 shillings.

Cheers

Don

[This message was edited by Don Atkinson on MONDAY 24 June 2002 at 18:58.]

I realised there might be a few football fans reading the last brain teaser and wondering who buys hogs these days. So I've tried to dress it up in a more relevant way. If it helps, you could substitute 'pints of larger' instead of 'hogs' or 'team shirts'. Remember, after two dozen pints, it might be useful to be able to do this brainteaser, just in case you've forgotten who your wife/girlfriend is!

Three of my mates who are fanatical football fans came to see me yesterday. They had just got back from the World Cup in Japan where they had been for past month, with their wives/girlfriends joining them last Thursday, just in time to see the England /Brazil game. The men's names were Henry, Colin and Chris. The wives/girlfriends were Georgie, Carol and Anna, but I forget the name of each man's wife/girlfriend.

They told me they had bought replica team shirts. Each person had bought as many team shirts as they gave shillings for each shirt (paid for in Yen/dollars/euro/lire/Rials etc depending on which shirts they bought). Henry bought 23 shirts more than Carol and Colin bought 11 more than Georgie. Likewise, each man laid out 3 guineas more than his wife/girlfriend. What was the name of each man's wife/girlfriend.

Note, a guinea is 21 shillings.

Sorry, I couldn't do much about the shillings/guineas !!

Cheers

Don

Vuk, a nice solution and slightly different method to mine, (but the same answer!)

I'll wait a day or two before publishing, but I am not claiming mine is any more elegant, just slightly different.

BTW, Vuk, is there a small typo in your line (Y+23)+(X+11)+Z=X+Y+Z+189
eg perhaps the first 'Z' should be 'C' ?

Cheers

Don

Vuk

Shouldn't that be

A^2+B^2+C^2=X^2+Y^2+Z^2+189?

[spent as many shillings on each hog as they bought hogs]

Matthew

The difference of two squares equalling 63 gives the following possible combinations for husband and wife. Diffrences of 1,3 and 7 are the only possibles giving integers.

32 and 31
12 and 9
8 and 1

Hendrick has 23 more then Catrinn (32 and 9)
Claas has 11 more then Geertrick (12 and 1)
Leaving Anna with 31 and Cornelius with 8

Hendrick and Anna
Claas and Catrinn
Cornelius and Geertrick

Matthew

Yesterday, Mrs Don over-heard me talking about the stunningly accurate clocks inside naim's cd players. Straight away she wanted to know whether it was a Rolex or an Omega and suggested, in no uncertain terms that it would be more usefull wrapped around her wrist than hidden in a black box on a shelf!

I muttered something about it being wrapped around her throat if she wasn't careful, and just managed to save the rapidly developing 'situation' by suggesting I had really meant 'neck' rather than 'throat' and preferably built into a diamond studded necklace to enance the radiant counternace of her delighful features etc etc......

I think I might have got away with it, because Mrs Don, who is a reception class teacher at one of the local girl's schools, then turned my attention to the kitchen clock. Now there is nothing special about the kitchen clock. Its a bog standard 12 hour analogue clock with a battery and an hour hand and a minute hand. The dial is marked off in 1 minute intervals with the usual 5, 10, 15 ..... marks highlighted and numbered 1, 2, 3.....etc.

Now Mrs Don, remembering her day at school with the five year olds and 'telling the time' lesson, then asked me,

"At what time(s) are the hour hand and minute hand exactly one minute appart AND at the same time, both pointing exactly at minute marks ?"

Help !!

Cheers

Don

DON

I think there are two answers:-

12 minutes past 2, and 48 minutes past 9.


STEVE D

quote:

---

"At what time(s) are the hour hand and minute hand exactly one minute appart AND at the same time, both pointing exactly at minute marks ?"

---

Sorry misunderstood the question at first, 2:12 and 9:48 are correct.
John

I have just arrived home (london) to a warm sunny evening after my holiday and am resetting my alarm clock and notice that the hands of both my watch and clock are directly opposite each other. I know my alarm clock is running fast, 2 and 3/4 minutes.

Where have I been?

Matthew

Don,

My pleasure at your recent visit to the World Cup thread is somewhat tempered by a sneaking suspicion that it might be part of some sophisticated plot to restore Brain Teaser to its former #1 status. Naturally as a football grunt such subtlety is way beyond the capacity of my modest brain.

Matthew

<Plants small bomb in corner of thread and sneaks off back to World Cup with an evil laugh>

Drat! I fell for it too.

Alex

<new maths puzzle hidden in footie thread rumour>

Matthew T

You must have been somewhere either 5 and a half hours ahead, or 6 and a half hours behind.

I can only find India or Nepal to satisfy this criterion.

STEVE D

Thought I had better post my solution to the 'hogs' teaser. Its not at all sophisticated.

Since each person bought 'as many hogs as they gave shillings for each hog' they each spent a SQUARE NUMBER of shillings. (Matthew T also noted this)

Also, the difference between the square numbers of shillings spent, has to be 63 for each couple.

We need to express 63 as the difference between two squares in three different ways viz,

(8^2 - 1^2); (12^2 - 9^2); (32^2 - 31^2)

I simply used a bit of trial and error ! and the requirement to have two of the numbers differ by 23 and another two differ by 11, but not necessarily in the same pairs.

The 32 and 9 are clearly Hendrick and Catriin and the 12 and 1 are Class and Geertrick.

Anna and Cornelius simply slot into the remaining places.

OK the 'bit of trial and error' was QUITE a bit of trial and error. I just couldn't see a shortcut.

Any advice???

Cheers

Don

Steved,

You are correct, I have been in India but it could have been Nepal

Don,

I used the a relatively straight forward method of evaluating the difference of two squares.

x^2-(x-n)^2 = 2nx-n^2

Equate this to 63 and find integers of n that give integer solutions for x.

Matthew

Answer: 0.000000001 (approx)

I think we can now safely reduce that to '0'

Strange how nobody else felt confident enough to put a figure to the probability of Germany winning..........

Cheers

Don

I have been in India but

I used to travel to India a lot. Mainly Bombay and Delhi but also Shrinagar, Puna, Bangalore, Marmagoa and Karwa. Beautiful country but heartbreaking poverty.

Cheers

Don

I noticed something interesting the other day……and I also noticed the following!

Write down ANY positive integer. eg 473947281
Now write down the same set of digits, but in ANY other order. You COULD just swap two digits or you could re-arrange the lot.......or any combination in between. eg 737824914.

Subtract the smaller from the larger eg 737824914 - 473947281 = 263877633

Notice anything interesting about the result, ie the 263877633 bit?

Cheers

Don

Divisible by 9.

Cool.

Paul

Because the difference between two arbitrary powers of 10 is divisible by 9.

10^m - 10^n = 10^n(10^(m-n) - 1), which is either 10^n times 9, 99, 999, 9999 etc.

(In Don's example all his numbers are divisible by nine, I'm sure that's unintentional...)

Paul

(In Don's example all his numbers are divisible by nine, I'm sure that's unintentional...)

I should have avoided examples with factors of 9 even though there was only a 1 in 9 chance of picking such numbers at random anyway.

Cheers

Don

I work as a flying instructor and spend most of my time in light, single-engine aeroplanes such as Cessnas and Pipers. Some of my time is spent in light twin-engine aeroplanes. Now most of the time, I don't worry about engine failures, first because they are pretty rare and second because we all spend a lot of time practising what to do in the event of an engine failure. So you can imagine my surprise, when glancing through a maths book today (very low cloud, no flying!) to see a problem based on engine failure rates of 1 in 100 flights (touch wood rapidly). OK the problem was set in the year 1900 ie before the Wright Brothers got off the ground, so to speak, but a 1 in 100 chance of a forced landing due to engine failure, phew!

Anyway, the general gist of the book was that IF we fit two engines, when only one is needed to keep aloft, and IF engine failures are independent (see note 1 below), then we now have reduced our chances of a forced landing to 1 in 10,000.

Now the pilock who was writing this book, reckoned that it would also be a good idea to continue this trend of having one 'redundant' engine. He therefore suggested using four engines when only three were needed to keep aloft. Assuming engine failures ARE independent events, what now would be the probability of an engine failure induced forced landing, or ditching?

Cheers

Don

Note 1. Engine failures aren’t necessarily independent, eg ones that have a common design fault or were produced in the same defective batch or were recently serviced by the same trainee mechanic etc etc

Sorry, i've been away this week.

Neat method. The other way is, of course, to recognise there are 6 ways in which two out of four engines could fail, plus the risk of three engines failing, plus the risk of all four giving up together. Of course the risks of three or four engines failing are much less than two engines and can therefore be ignored (unless you're the driver or a passsenger!)

So the chances for failure are 1-(0.99+0.04)(0.99^3), which is 0.00059203

Which approximates to 1 in 1680 flights which is still a bit too risky. Also puts common phrases such as "9 times out of 10", "99 times out of a hundred" or "99.9% of the time..." into some sort of context ie "= not good enough!"

Cheers

Don

I have £10k in £10 notes stashed away in 10 bags.

My local Naim dealer sells hifi stuff, priced from £10 (for a bit of 'green stuff') to £10k for a 552 preamp and he has something at every £10 price-point in between.

I have just checked my 10 money bags and realised that I could use a whole number of bags to buy ANY single item in the store, (without the need for change)

How much money do I have in each of my bags ?

Cheers

Don

Omer,

10 + 20 + 40 + 80.... + 5120 = >£10k

I only have £10K !!

But I can see where you're comming from, and yes, I know I'm nit-picking, but elsewhere others are complaining about the quality and accuracy of threads................

Cheers

Don