Brain Teaser No 1

Dan,

Keep going. You are getting there.

Your estimate of Paul D's original salary would be £1,080 at the moment (give or take £9)

Believe me, he earns more than that.

But NOT a lot more

Poor chap!

Cheers

Don

Paul,

Thanks for hosting the probability tree.

It looks ok to me, but then it would, wouldn't it....

It allows you to EASILY find the probability of (say) 3xReds (1/220) etc etc

Many thanks

Don

I was just getting worried that there may have been some goats in there somewhere

No chance....

Cheers

Don

No chance....well, if you play your cards right, it could be as high as 2/3....but we don't want to start that debate again, or do we?

quote:

---

Originally posted by Don Atkinson:

Your estimate of Paul D's original salary would be £1,080 at the moment (give or take £9)

Believe me, he earns more than that.

But NOT a lot more

Poor chap!

---

Don,

I believe the answer is £1,350. I am keen to see your solution/method.

cheers,

Dan

[This message was edited by Dan M on Fri 22 October 2004 at 0:16.]

I believe the answer is £1,350.

Dan, You've earned that big fat cigar!!!

Well done.

I am keen to see your solution/method

I have an elegant little solution, but the size limit for posts on this forum are too narrow.....

Cheers

Don

PS, I will type up my method over the weekend to post it

quote:

---

Originally posted by Don Atkinson:
I have an elegant little solution...

---

Ah good! Mine resorted to brute force - not elegant at all. I look forward to seeing your method.

-Dan

If Don isn't the only one allowed to ask questions, a colleague asked me this the other day and said very few people ever got the answer. I got it after a while and to my surprise the other people who I have now asked the question have failed to get it, so:

Bags of coins

You have been given 10 bags of coins, in each bag there are 10 coins (100 coins in total). All of the coins look identical. Each of the 10 coins in 9 of the bags weigh 10 ounces each and in 1 of the bags each of its 10 coins weigh 11 ounces each.

You have also been given a set of scales, you have to identify which bag has the 11 ounce coins in it by making only 1 measurement on the scales.

If Don isn't the only one allowed to ask questions,

Don DEFINITLY isn't the only one allowed to ask questions......

Its great when others ask them, and they have a habit of asking the best ones.....yours looks like one of these.

Are the scales of the direct reading type, or are they of the old fashioned "scales of justice" type ie with two pans and no dials?

Or doesn't it matter?

Cheers

Don

Are you allowed to remove coins from the bags before weighing?

Cheers

Don

Magic Squares

The elements of any 3x3 magic square can be represented with three parameters e, x, and y as shown in the matrix below (note the symmetry)

(e + x)...........(e - x - y)..........(e + y)

(e - x + y)..........e.............(e + x - y)

(e - y)...........(e + x + y)..........(e - x)


These elements can be rearranged into a magic square where rows are arithmetic progressions with a common difference (x) and the columns also have a common difference (y). (note the symmetry)

(e - x - y).........(e - y).......(e + x - y)

(e - x)...............e...............(e + x)

(e - x + y).........(e + y).......(e + x + y)

We need to find a magic square with primes [ P] which have the property that [P+2] are also primes.

The next bit is the "neat" bit......although it it basically trial and error

Unless 3 or 5 appears as an element, the [P] terms must all terminate in 9, 7 or 1.
The e must be the prime mean of two primes ending in 9, 9; or 7, 1; or 7, 7; or 1, 1 respectively.

Next, take a look at the sequence formed by the smaller members of the successive twin primes (ie the primes that occur in pairs with a difference of 2)

3, 5, 11, 17, 29, 41, 59, 71, 101, 107, 137, 149, 179, 191, 197, 227, 239, 269, 282 ........

We now need a bit more trial and error!!!!!

(for any prime < 149 and ending in 1 or 7 there are not more than three smaller primes in the sequence with the same terminal digit)

The smallest value of e ending in 9 and which is the mean of at least four prime pairs can be seen as

2(149) = 107 + 191 = 101 + 197 = 71 + 227 = 59 + 239 = 29 + 269 = 17 + 281

So we put 149 into the centre, then successively fit (= that famous trial and error technique again !) the above pairs into the middle row of the arithmetic progression array, there is only one arrangement that allows us to fit three of the other pairs properly


17....59....101

107...149....191

197...239....281

From which we can soon develop the three required magic squares

191....17....239...............192.....18.....240...............193......19....241

197...149....101...............198.....150....102...............199.....151....103

59....281....107................60.....282....108................61......283...109

So Paul D's salary was 9e + 9 = £1350

But has now shot up to the astonishing sum of £ 2700

......or has it ?

There is another answer......

MUCH bigger than 1350....

Just when you thought it was safe to go back to sleep.....

Cheers

Don

Dan,,
That was my elegant little solution.....

Not much better than the old "trial and error" method really.

Cheers

Don

Bags of coins

Don asked a couple of questions:

Are you allowed to remove coins from the bags before weighing?

Are the scales of the direct reading type, or are they of the old fashioned "scales of justice" type ie with two pans and no dials?

Yes you are allowed to take coins out of the bag.

The type of scales does not matter. You can put one set of items on the scales and take a reading and that is it.

Bags of Gold...

You can put one set of items on the scales and take a reading and that is it

So, I take 1 coin out of the first bag, 2 coins out of the second bag, 3 coins out of the third bag etc.

I then put these 55 loose coins onto the scales and weigh them..

If the weight is 550 ozs then you were cheating on all the coins weighed 10 ozs each

If the weight is 551 ozs the first bag is the odd one out

If the weight is 552 ozs the second bag is the odd one out

If the weight is 553 ozs the third bag is the odd one out

etc

Any good ?

Cheers

Don

Two-Sheds

Your teaser is a GENUINE teaser, not a thinly diguised A-Level maths question.

It should have also appealed to Chris Patterson and I am deeply disappointed he didn't take part......

Cheers

Don

PS don't tell me my answer is wrong, that would be deeply embarrasing....

quote:

---

you have to identify which bag has the 11 ounce coins in it by making only 1 measurement on the scales.

---

Say you were to label the bags A through J and then take 1 coin from A, two from B etc up to 10 from J. Then weigh all the coins you've collected together.

If every bag contained 10oz coins we'd have 55oz. If A has the big coins then we get 56, B, 57 through to J giving 65.

So one measurement tells us which bag contains the over-weight coins. Unfortunately unless we've been busy with the marker pen on each coin we have no way to put them back in the correct bags without further measurement...

Edit: Blimey, Don's post wasn't there when I started... And I seem to have lost an order of magnitude's oz. Never mind.

Paul

Paul,

Unfortunately unless we've been busy with the marker pen

you could keep the coins in NEAT and SEPARATE piles on the scales.........

Cheers

Don

well done Don and Paul you are indeed both correct.

Nice one, Two-Sheds

Any more?

Cheers

Don

Looking back at the Matthew and Mick Bike teaser, I realised that I haven't posted my "Trial & Error" solution. So here goes

I created a tabular format to keep tabs on the numbers and I calculated the speed, average speed and distance for each cyclist at 1 second intervals. Dimensions are secs, m/s and m. The green light is at (time 0) and (dist 0) for Matthew and at (time 4) and (dist 0) for Mick.

As if by magic, most of the numbers are whole digits and the time at which Mick passes Matthew and the time at which Matthew passes Mick are nicely on whole seconds.

…………Matthew…………….……………..Mick
Time…..Speed…..S(avg)……..Dist….……..Speed…….S(avg)……….Dist
0………..1……………………..0
……………………2
1……….3………………………2
……………………4
2……….5………………………6
……………………6
3……….7……………………...12
……………………8
4……….9………………………20………….16…………………………0
…………………..10………………………………………16.5
5………11……………………..30…………..17…………………………16.5
…………………..12……………………………………….17.5
6………13……………………..42…………...18…………………………34
…………………..14……………………………………….18.5
7………15…………………….56……………19…………………………52.5
…………………..16……………………………………….19.5
8………17…………………….72……………20…………………………72
…………………..18……………………………………….20.5
9………19…………………….90……………21…………………………92.5
…………………..20……………………………………….21.5
10……..21……………………110…………...22…………………………114
…………………..22……………………………………….22.5
11……..23……………………132……………23………………………...136.5
…………………..24……………………………………….23.5
12……..25……………………156……………24………………………..160
…………………..26……………………………………….24.5
13……..27……………………182……………25………………………..184.5
…………………..28……………………………………….25.5
14……..28……………………210……………26………………………..210


Of course, there is no guarantee that this table will copy neatly into the forum. Hope.

The table shows that Mick passes Matthew 72m from the light. This happens 8 secs after Matthew has passed the light. At this time Matthew is doing 17m/s and Mick 20m/s

The "dist" is got simply by adding up all the preceding average speeds. Eg after 8 secs Matthew has travelled (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16) = 72. In practice its even easier because all you do is add 16 to the previous distance of 56…..

The next part of the table shows Matthew passing Mick 210m from the light. This happens 14 secs after Matthew has passed the light. At this time Matthew is doing 29m/s and Mick is doing 26m/s.

Of course, there is no guarantee that this table will copy neatly into the forum. Hope.

Oh dear....

Everything is misaligned.

The columns are labled
Time/Speed/Average Speed/Distance/Speed/Average Speed/Distance

The time is common
The next 3 columns relate to Matthew and last 3 to Mick

The figures aren't too bad, but they do need to be spaced out across the page to line up with the headings

I hope anybody really interested will be able to follow the numbers

You can always cut and paste them back into Word and get them lined up....

Cheers

Don

If you wrap a table in 'code' tags it comes out in a constant width font, I've no idea what this is supposed to look like though,

         Matthew                   Mick
Time   Speed   S(avg)   Dist     Speed   S(avg)   Dist
 0       0       0        0        0        0       0
 1       0       0        0        0        0       0
 2       0       0        0        0        0       0
 3       0       0        0        0        0       0
 4       0       0        0        0        0       0
 5       0       0        0        0        0       0
 6       0       0        0        0        0       0
 7       0       0        0        0        0       0
 8       0       0        0        0        0       0
 9       0       0        0        0        0       0
10       0       0        0        0        0       0

etc.

Paul

If you wrap a table in 'code' tags it comes out in a constant width font, I've no idea what this is supposed to look like though,

If you wrap a table in 'code' tags it comes out in a constant width font, I've no idea what you are talking about though. However, it looks superb and is just what I needed to know to get my table right.

So what are 'code' tags?? and how do you get such a neat table??

Remember, I know 'bugger' all about computer-speak

What I do know is that I haven't a clue how to format things to post on this forum and the loss of that little paper-clip is a disaster

But i'm willing to try new tricks. many thanks Paul

Cheers

Don

If you go to the full 'Post a Reply' window and press the rightmost button with '</>' on it you'll get some 'code' tags. Just like a 'quote' or even 'i' for 'italic'. With 'code' everything between the tags will retain its formatting and be rendered in a constant pitch font. So you can lay stuff out with the space bar. Crude but better than nothing.

Paul

Paul,

Many thanks

Will try again soon

Chers

Don