Brain Teaser No 1

quote:

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Can you write the numbers 1 to 9 in such a sequence such that it isn't possible to extract from the sequence ANY four-digit rising or falling subsequence?

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can I just say no?

if you start with 1 then the digits 2, 3 and 4 have to come somewhere in the sequence so you will have a rising sequence. Likewise if I start with 9 the 8, 7 and 6 have to follw somewhere.

The only digit I can start with where there is no four digit rising sequence is 7 since there can only follow 8 and 9 and hence only a 3 digit rising sequence, but 6, 5 and 4 have to follow so we have a falling sequence.

if you start with 1 then the digits 2, 3 and 4 have to come somewhere in the sequence

Correct.

BUT the 2, 3, 4 don't HAVE to come in that order, and providing (say) the 2 and the 3 were reversed, the subsequence would become 1 3 2 4 which isn't a rising or falling sequence.

Does this help ?

Cheers

Don

how about this then...

3 2 7 5 1 4 9 8 6

how about this then...

Excellent (9/10)

Why not "perfect" (10/10) ?

Well, I have seen a more "elegant" solution.....

Cheers

Don

outstanding brain teasers

No, not outstanding in the sense of being "superb", just in the sense of "not yet attempted".

Commuters on a platform (page 96)
Paul D's salary (page 96)
Matthew & Mick on bikes (above)
Three AV boxes (above)

Only Paul D's salary should give rise to any concern....

Cheers

Don

Matthew & Mick go cycling together....

Mick will overtake Matthew 96m after the lights, but matthew will have revenge going past mick (assuming no more traffic lights) 224m after the lights.

First thing is see what each of them is doing after 4 seconds.

Matthew:
first using the equation v = u + at (velocity = initial speed + (acceleration x time)
v = 0 + (4x2)
v = 8m/s
then d = ut + at^2 (distance = (init speed x time) + (accel x time squared)
d = 0x4 + 2x4^2
d=32m

Mick
v = u + at
v = 16 + (4 x 1)
v = 20

so when Mick passes the light Matthew is doing 8m/s and is 32m passed the light, Mick will be doing 20m/s and is at the lights.

for Matthew distance from light:
d = ut + at^2 + init dist from light
d = 8t + 2t^2 + 32

Mick
d = ut + at^2 + init dist from light
d = 20t + t^2 + 0

so when Mick Passes Matthew d must be equal:
8t + 2t^2 + 32 = 20t + t^2
0 = -t^2 + 12t - 32

using the equation for quadratics:
if a x^2+ b x + c = 0 then:
x = (-b +- sqrt(b^2 - 4ac))/2a

this gives t = 4 or 8

so plugging this into matthew's equation for distance past light:
d = 8t + 2t^2 + 32
d = 96m or 224m

Two-sheds,

What I am about to say is dreadful......

You get FULL marks for trying, and for taking part, and for being able to remember and use the old v=d/t type of equations......

BUT (oh I don't really want to do this....)

Mick........ passed the green light 4 seconds after Matthew at 16m per sec,

....somehow you have got this down as 20m per sec......

Can I persaude you to have another look?

For those of you who can't recall the equations that two-sheds has used, this problem lends itself rather nicely to a bit of trial and error.

Cheers

Don

PS Two-sheds DOES get the *bonus grade for realising that Matthew gets his revenge further along the road....but again the actual figure needs adjusting !!

quote:

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You get FULL marks for trying, and for taking part, and for being able to remember and use the old v=d/t type of equations......

BUT (oh I don't really want to do this....)

Mick........ passed the green light 4 seconds after Matthew at 16m per sec,

....somehow you have got this down as 20m per sec......

Can I persaude you to have another look?

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I think I mis-read your question (because your question is obviously crystal clear )

you said:

quote:

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Mick has been approaching the same lights at a steady speed, but on recognising Matthew has put on a bit of a dash, and passed the green light 4 seconds after Matthew at 16m per sec, accelerating at 1m per sec per sec.

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I read that to be Mick saw Matthew at the lights and started accelerating immediately, so he had been accelerating for 4 seconds before passing the lights, hence when he passes the lights he has already accelerated up to 20m/s. If however he only bothers to start accelerating at the light then...

(plus I made some other boo-boo's, intential of course to see if everyone was paying attention )

Matthew, velocity and siatance after 4 secs (I misses matthews init v of 1m/s):
v = u + at
v = 1 + 2x4
v = 9

(second boo-boo wrong formula for distance)
d = ut + 1/2 x at^2
d = 1x4 + 1/2 x 2 x 4^2
d = 20

As Mick passes the light, for matthew
d = ut + 1/2 x at^2 + init dist from light
d = 9t + 1/2(2t^2) + 20
d = t^2 + 9t + 20

Mick
d = ut + at^2 + init dist from light
d = 16t + 1/2(t^2) + 0

so when Mick Passes Matthew d must be equal:
t^2 + 9t + 20 = 16t + 1/2(t^2)
0 = -1/2 (t^2) + 7t - 20

using the equation for quadratics:
if a x^2 + bx + c = 0 then:
x = (-b +- sqrt(b^2 - 4ac))/2a

t= (-7 +- sqrt(7^2 - 40))/-1
t = (-7 +- 3) / -1

this gives t = 4 or 10

so plugging this into matthew's equation for distance past light:
d = t^2 + 9t + 20
d = 72m or 210m

so mick passes matthew after 72m and matthew passes mick after 210m...

hopefully I;ve paid more attention this time...

Two-sheds

hopefully I;ve paid more attention this time...

Spot on. The answers are now perfect

Cheers

Don

BTW, How fast would each be going when they pass each time?

speeds when they pass each other.

v = u + at

when Mick passes Matthew
Matthew:
v = 9 + 2 * 4
v = 17m/s
Mick:
v = 16 + 4
v = 20m/s

when Matthew passes Mick
Matthew:
v = 9 + 2 * 10
v = 29m/s
Mick:
v = 16 + 10
v = 26m/s

Boxes for Heathrow

The total number of combinations of boxes (t) is (n = total number of boxes):
t = n * (n-1) * (n-2)
t = 12 * 11 * 10
t = 1320

The total number of combinations that will result in one of each (q) is:
q = num of dvd's * num of av2s * num of nap175s
q = 3 * 4 * 5
q = 60

chances he got the right combination are
60/1320 = 1/22 so there is a 1 in 22 chance that he gets one of each.

Benford's Law

On page 91 Redgirl73 asked if someone could explain in simple terms just what Benford's Law was.

Can anyone explain what exactly is Benford's Law in layman's terms, as I've never heard of it!

At that time I couldn't. Now I think I can, so I hope Redgirl73 still looks in from time to time.

Benford observed that tables of numbers had a much higher proportion of numbers starting with 1, 2 or 3 than would be expected. He had expected numbers to start with each of the 9 digits with equal probability ie around about 0.11. But he noticed that numbers starting with 1 occurred much more frequently, with a probability of about 0.3.

He found a simple formula to describe his probability distribution with reasonable accuracy for many lists of numbers. The formula was (Logarithms to Base 10)

P = Log (1 + 1/D)

Where P = probability of a number stating with D

He also proposed that the probability of a number starting with (say) 1, 5, 8 (in that sequence)

P = Log (1 + 1/[158])

How do I know? I read it in a book (but don't believe everything you read in books!)

Cheers

Don

Two-Sheds

I haven't checked the speeds yet, but 26m/s is about 94 kph, which is going some, even for Matthew......

Cheers

Don

PS I will post my "trial & error" solution in the next day or so, along with the answer (ie whther you are right or wrong)to the av system in the warehouse.

Magic squares

Don,

Interesting teaser. I haven't come aross 'magic squares' before, but a quick google implies the numbers are sequential and start at 1, which doesn't fit with your problem. So, for my sake, can you define one. I'm assuming the numbers must be unique, yes?

Dan

Magic Squares

Dan,

Thanks for picking this up and asking about magic squares.

We are talking about a 3 x 3 square in this teaser. The letters a to i could represent a magic square. The letters are arranged in three rows each with 3 letters like so

a b c

d e f

g h i

For the square to be a MAGIC square, each line of three numbers must add up to the same answer

So, a+b+c = a+d+g = a+e+i etc

Does this help? if not, just say so, and I will try again.

Good Luck

Cheers

Don

Clean-up time for a couple of outstanding teasers.....

Number flushes

Can you write the numbers 1 to 9 in such a sequence such that it isn't possible to extract from the sequence ANY four-digit rising or falling subsequence?

eg 1 2 3 5 4 6 9 7 8 ISN'T acceptable because 1 2 3 5 is a rising subsequence as is 1 2 3 4 and 1 2 3 9 etc etc. But there are no four-digit falling sub-sequences in this particular sequence.



Two-Sheds said

how about this then...

3 2 7 5 1 4 9 8 6

Which is a good answer.



I suggested a (slightly) more elegant solution existed

Well, in fact there are several eg

3 2 1 6 5 4 9 8 7

ie three, triple reverse runs

Cheers

Don

Three crucial boxes for Heathrow...

Don, I think you would have to show your working. Stats was never my strong point at school, but I think the approach I took is the right one.

Gentlemen,

You are both clearly heading in the right direction (IMHO)

(IMHO ?) well....I still haven't found the question/answer, so I am no more able to guarantee that my answer is right than you are.....

As for publishing my method, I have no problem, in principle, with doing that. I drew a "probability tree". You know, start at the left and draw more and more branches towards the right ?

However, let me first suggest three things....

Remember, the question asks for the probability of picking one ball of each colour.

If the first ball out of the bag is Blue.....you could still end up with one of each colour

How many ways could you pick R/W/B ?

Does this help ?

Does anybody want to change their answer? (I don't, well, at least, not yet !)

Cheers

Don

PS (added after posting the above)

I am drafting out an explanation of my method (not easy to post a diagram these days !) but I think we'll agree on the answer before i've finished and published it.

[This message was edited by Don Atkinson on Mon 18 October 2004 at 18:34.]

Don't think it would be right to add these probabilities.

Alex, I'm with you on that one !

1/4 + 4/11 + 5/10 > 1.0 !!!!

and 1.0 is a dead cert!!!

Cheers

Don

I'm with Don.

There are 12.11.10 selections possible of which (3.4.5)(3.2.1)meet the criteria. So 3/11.

If I try and explain in more detail it will just get more complicated. I'm sure Don's diagram will epitomise clarity. If he'd like me to host it my email is in my profile.

Edit... No it's not. paul at kilmory dot demon dot co dot uk

Paul

Paul,

Many thanks for offering to host my diagram.

Not my best, but I hope it is clear.

Cheers

Don

quote:

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Originally posted by Don Atkinson:
_Naim Double Paul Darwin's Salary....._

Now Paul D set to work immediately arranging his salary into nine unequal piles of coins, but triumphantly realised he had created a 3x3 'magic square'.

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I thought I had it for a minute:

30. 138 192 : 360
228 18. 60. : 306
102 150 108 : 360
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360 306 360

You'll notice that adding and subtracting 1 gives all primes as required. Of course the diagonals are way off. Oh well, back to the drawing board...

cheers,

Dan

Three crucial boxes for Heathrow...

Paul you said:

quote:

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There are 12.11.10 selections possible of which (3.4.5)(3.2.1)meet the criteria. So 3/11.

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as I said my stats was never that good, so could you explain the (3.4.5)(3.2.1) a bit more. I can see where you get (3.4.5) from, but where does (3.2.1) come from? and what are you doing with all those numbers to get 3/11?

Don, Paul - are these solutions saying that if you look and you have picked out a dvd and another dvd you don't care what the third item is since it's wrong?

Don's document is here. If the original 'doc' format proves unpopular I can have a go at transforming it.

If you pick three Naim boxes at random from the twelve in the storeroom you get one of 12.11.10 possible selections. This is treating the order as important. The probability that you've chosen an appropriate set of boxes is then the ratio of acceptable selections to the total possible. The number of acceptable selections is 3.4.5, but we don't care what order we pick them in, so we multiply by the number of ways of arranging 3 items, 3.2.1. The result is then 1.2.3.3.4.5/12.11.10, the 10 and the 12 cancel with the 2.5 and 3.4 leaving 3/11.

Paul

ok, I can see now that Master Don is correct from his diagram. I was just getting worried that there may have been some goats in there somewhere.

Still no cigar...

102 18 240 360
198 150 12 360
60 192 108 360
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360 360 360 360