Brain Teaser No 1

Well, God granted him to be a boy for the sixth part of his current life, and adding a twelfth part to this he called him a hansom youth (Fiction, pure fiction); After a further seventh part he took a partner in love and five years later was granted a son. Alas! the wretched child, after attaining the age of half his father's current life, joined Mana as their sales director. Four years have passed since that dreadful day but Paul's consolation has now been realised with the launch of Fraim. To celebrate this glorious event and his xxx birthday, Paul has just decided to deliver a beautiful 5 tier Fraim supporting a nicely run-in NAP500 to Don Atkinson......Ok, Ok, I SAID it was fiction, pure fiction......

So, according to the facts above, how old is Paul?? (Lets hope his sense of humour is just as great!)

Fiction, pure fiction so its no good ringing Naim, even Paul doesn't know how old he is - yet!

Cheers

Don

Onto weightier matters.

I was interested in whether the first principles argument leading to 1,3,9,27 produced a general minima for number of weights should inflation strike and the maximum weight need to be extended arbitrarily.

Boing.

Paul

Got 84. Twice.

Magic, answer is ok, although I could swear that Paul S didn't look a day over 40.

Would you like to publish the explanation/working-out for the benefit of those who are still struggling? Or prefer to wait and allow someone else that qudos?.

I realise this was an 'easy' one - well Duncan F asked for a few more that he could get his mind around! perhaps Duncan could help out?

Cheers

Don

quote:

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Would you like to publish the explanation/working-out for the benefit of those who are still struggling?

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Here goes!

You just write an equation, L is his current age, and put in everything you're told, so

quote:

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Well, God granted him to be a boy for the sixth part of his current life

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L/6 +

quote:

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and adding a twelfth part to this he called him a hansom youth

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L/12 +

quote:

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After a further seventh part he took a partner in love

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L/7 +

quote:

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and five years later was granted a son.

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5 +

quote:

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Alas! the wretched child, after attaining the age of half his father's current life,

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L/2 +

quote:

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joined Mana as their sales director. Four years have passed since that dreadful day but Paul's consolation has now been realised with the launch of Fraim

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+ 4 = L

So

L/6 + L/12 + L/7 + 5 + L/2 + 4 = L

or

9/12L + L/7 + 9 = L

25/28L + 9 = L

So 3/28L = 9, L = 3*28 = 84

Paul

1 + 3 + 3^2 + ... + 3^n = (3^n+1 + 1)/2

but perhaps it shouldn't be? Well out of practice at this maths stuff.

Of course 'we' haven't proved yet that there aren't alternative combinations of weights that are equally efficient somewhere between 1kg and infinity.

Paul

SN = r.(x^0+x^1...+x^N)

Multiply by x

x.SN = r.x.(x^0+x^1...+x^N)

or

x.SN = r.(x^1+x^2...+x^(N+1))

Subtract

x.SN - SN = r.(x^1+x^2...+x^(N+1)) - r.(x^0+x^1...+x^N)

or

SN.(x-1) = r.(x^(N+1) - x^0)

therefore

SN = r.(x^(N+1) - 1)/(x-1)

Matthew

quote:

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Sn = (q^n-1)/(q-1), where Sn is the sum of n first elements and q is the quotient (3 in this case). I guess this formula is easily proved using induction.

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Penny's are begining to drop. (A level maths was more than 20 years ago now, it is excusable...)

Sn = (q^n+1 + 1)/(q-1) (1)

So Sn+1 = (q^n+1 + 1)/(q-1) + q^n+1

= (q^n+1 + 1 + q.q^n+1 - q^n+1)/(q-1)

= (q^n+2 + 1)/(q-1)

So if (1) is ever true for any n, it is also true for n+1. QED....

Paul

quote:

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I realise this was an 'easy' one - well Duncan F asked for a few more that he could get his mind around! perhaps Duncan could help out?

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I was bashing away using my patented solution finding method of "proof by exhaustive writing down" when I stumbled on Paul's magic number of 84. I think I'd got as far as noticing that 6, 7 and 12 were common factors of 84 which led me to to believe that I could show 84 to be an answer, but with no idea of whether or not alternative solutions were possible

Paul's explanation seems much more elegant!

[This message was edited by Duncan Fullerton on TUESDAY 08 January 2002 at 17:33.]

I was bashing away using my patented solution finding method of "proof by exhaustive writing down"

Sounds very familiar to me. I am still bashing away like this on Bam's teaser about the ladder.

I now recal from school days that if the roots of ax^2 + bx + c = 0 are Alpha (A) and Beta (B), then

A + B = -b/a

AB = c/a

Also if B = 1/A then a = c (or conversely if a = c then the roots are probably reciprocals)

Now i don't know if any of this will help me, but I am bashing away using my patented solution finding method of "proof by exhaustive writing down"

Thanks Duncan,

Cheers

Don

t = D/v1 + D/(v2-v1)

A bit of twiddling gives,

t = D(v2/(v1v2-v1v1))

D is a random constant, for convenience v1 is 1, so,

t 'is proportional to' v2/(v2-1)

This looks plausible, if v2 == v1 then it takes forever for the troopers to meet. The conclusion is that the faster v2 is relative to v1 the shorter the time to meeting.

I wonder....

Paul

[This message was edited by Paul Ranson on TUESDAY 08 January 2002 at 23:14.]

"WALK south until you find your mate's 'chute, then RUN like hell to catch him up"

Assumming they land 1km appart and walk at 1m/s and run at 4m/s, they would meet after 1,333 secs, at a point 1,333m south of the southerly parachute. This is purely for illustration purposes only.

But I have a better idea!

Cheers

Don

Details will follow when I have time to think this idea through!

However, the idea is that the paras run (or walk) 100m north; 200m south; 300m north; 400m south etc until they either meet the other para or find the other 'chute.

IF they both set of at the same time and in opposite directions (very unlikely) they will meet soonest. IF they set of in the same directions it will probably take longer and they will find the other 'chute first. (I think) in which case they then continue towards the other guy whom they will eventually meet on his next leg towards that 'chute.

As I say, just an idea and I need to give it a bit of thought!!

Cheers

Don

Evaluating relative performance seems like a job for a simulator.

Paul

quote:

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Originally posted by Don Atkinson:
However, the _idea_ is that the paras run (or walk) 100m north; 200m south; 300m north; 400m south etc until they either meet the other para or find the other 'chute.

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You're going to be passing your own 'chute every time you switch back. Going to be fun coding the if statement that makes you ignore this one but still stop when you find the other one.

cheers, Martin

I know this isn't quite the original problem, but anyway...

- GregB

Insert Witty Signature Line Here

We can tell our parachute from our chums because we know where we are. We don't check for a parachute when we pass ours.

Paul

Paul

Your pre-Xmas ladder teaser still hasn't been solved. So I gave it a bit more thought today! For 'thought' read Duncan F's method of 'proof by exhaustive writing down', in the hope that something works - BTW, I thought this was previously called serendipity??

First, I must come clean, and point out that I still haven't got, what I would describe, an ELEGANT solution. However, since your challenge was to get the MOST elegant solution, then to date I must claim the prize, only because nobody else has published ANYTHING, never mind anything elegant. However, this would be a hollow victory, to say the least!

Now I must also confess that I spent some time looking for a SIMPLE practical solution; one that can be done with pencil and paper on a dark December night, without the need to find square roots. Although WHY, I don't know, because finding square roots is no more difficult than long division!

So I offer, purely as an INTERIM measure, two solutions. The first,(which I mentioned a few pages back!) I think, is sort-of elegant, it is precise, but involves finding square roots. The second is simple, reasonably accurate and avoids square roots, but requires a bit of simple iteration for ladders less than (say) 5m long.

I will actually publish them shortly after this post, and probably as image attachments.

Of course, the real challenge ie finding some elegant (and most likely bloody impractical) solution involving pi; e, logs, the sqrt of -1, and the sum of an infinite converging geometric series, lies hidden from our minds, (well mine anyway) just below the surface!!! much to Bam's great amusement!! Although I doubt if the Sony engineers remembered to bring their 4 figure tables with them!

Cheers

Don

I hope there aren't too many typos. If its not easily legible, e-mail me and I will send a copy in Word.

Marks out of 10 Bam ?

Cheers

Don

Cheers

Don

Well, this didn't look any better so as a stop-gap I will copy the text below, but of course the 'superscript' and 'subscript' will be lost! I have inserted some ^s to show squares but I can't sort underlining to show "all divided by 2". Hope it helps!

Cube of side 1
L = Length of ladder
d = Distance to base of ladder from the cube

Height of ladder above cube = 1/d (Similar triangles)
L^2 = (1+d)^2 + (1+1/d)^2 (Pythagoras)
L^2 = d^2 + 2d + 2 + 2/d + 1/d^2 (1)

Let y = (d + 1/d) (2)
hence y^2 = d^2 + 2 + 1/d^2 (3)

Sub in (1) for y and y^2
L^2 = y^2 + 2y
y^2 + 2y - L^2 = 0

y = -2 +- (4 + 4L^2)^0.5
2

y = -1 + (1 + L^2)^0.5 (we're only interested in the + root)
y = (L^2 + 1)^0.5 - 1 (4) (This is the first equation to calculate)

y = d + 1/d (5) (From (2) above)
d^2 - yd + 1 = 0 (Multiply both sides of (5) by d and re-arrange)

d = y - (y^2 - 4)^0.5 (6) (sub value of y from equ (4) and solve for d)
2 (note; we're only interested in the - root)

Note: The ONLY equations the Sony engineers actually need are (4) and (6) and the arithmetic is quite simple. I think this is sort-of elegant and it gives precise answers!

[This message was edited by Don Atkinson on SUNDAY 13 January 2002 at 23:16.]

[This message was edited by Don Atkinson on SUNDAY 13 January 2002 at 23:17.]

[This message was edited by Don Atkinson on SUNDAY 13 January 2002 at 23:19.]

[This message was edited by Don Atkinson on SUNDAY 13 January 2002 at 23:20.]

I will award 8/10 for a single polynomial-style equation relating d to L. I will award 10/10 and a bottle of Moet & Chandon to the provider of the other, more elegant way of solving this.

BAM
(You've been very quiet on this one, Omer)

8/10?

Matthew

The second solution, which requires a bit of iteration, but avoids finding square roots is attached.

Again, because I haven't sorted out pictures/images, I am also pasteing a copy below.

Note, the main diffrence to solution 1 is the use of the approximation that

(A^2 + B^2)^0.5 = A + (1/2A) - (1/4A)^2 + (1/8A)^3 - (1/16A)^4

Cheers

Don

Cube of side 1
L = Length of ladder
d = Distance to base of ladder from the cube

Height of ladder above cube = 1/d (Similar triangles)
L^2 = (1+d)^2 + (1+1/d)^2 (Pythagoras)
L^2 = d^2 + 2d + 2 + 2/d + 1/d^2 (1)

Let y = (d + 1/d) (2)
hence y^2 = d^2 + 2 + 1/d^2 (3)

Sub in (1) for y and y^2
L^2 = y^2 + 2y
y^2 + 2y - L^2 = 0

y = -2 +- (4 + 4L^2)^0.5
2

y = -1 + (1 + L^2)^0.5 (we're only interested in the + root)
y = (L^2 + 1)^0.5 - 1
y = L - 1+ 1/2L - 1/(4L)^2 + 1/(8L)^3 - 1/(16L)^4 (Approximately) (4)
(This is the first equation to calculate)

y = d + 1/d (5) (From (2) above)

Now, rather than using the famous formula 'y = -b +- sqrt(b^2 - 4ac) all over 2a',
for any value of y, you can have a pretty good guess at a value of d just by 'inspection'

For example, IF y = 10, a pretty good guess for d and 1/d would be 9.9 and 0.1.
and IF y = 4, a pretty good guess for d and 1/d would be 3.75 and 0.25

Now each one needs a bit of iteration (jiggling!) to refine the precise values of d and 1/d remembering that the answers eventually have to add up to y AND be exact reciprocals. Not elegant, but saves having to find square roots!

Now I know this will get me 8/10 because it is a single equation. It's cheating, of course. The two equations that I posted above, and which I have had on my notepad since Xmas, can easily be combined into this equation. Of course, its then completely impossible to describe the equation to anybody. Which is why I didn't do it before now!

d = [√(L^2 + 1) - 1 ± √{L^2 - 2(√(L^2 + 1) + 1)}]/2

Now, at the moment, I can't see a way to simplify this equation. It should be possible. Sooner or later the penny will drop. I know that (L^2 + 1)^0.5 can be approximated to l + 1/(2L) etc although my recollection of binomial expansions seems to stop at positive integer polynomials for the moment!

Perhaps I will try other forms of pythag such as Sin^2(A) + Cos^2(A) = 1 etc and see where that leads.

Nice little teaser Bam!

Cheers

Don