Brain Teaser No 1

I light at the same time both ends of one stick and the one ned of another. When the stick with both ends lit is burnt out I light the end of the other stick. When that stick has burnt out 45 minutes have passed.

Matthew

b) the probability that a man who owned all of these items would be innocent? must be about 1in 10 ie pretty high

It's actually 9 in 10, but you get the idea.
John

quote:

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Originally posted by Vuk's son:
1, -1, 2 ?

No direct way, I just thought that in order for the the 3rd formula to be integer, two of the three numbers may cancel each other hence x=-y.

So trying x=-y, from (1) we get z=2, and from (2) we get x=1, y=-1, it fits.

Omer

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you have the right solution, omer, but i am looking for a general method that will work for ANY consistent right hand side, which may, or may not be all integer.

i like your reasoning though.

any other contributions?

enjoy

ken

any more "offers" on the non-linear simultaneous equations? or is this one of those that you are reserving for after-christmas-dinner?

enjoy

ken

Ken,

Interesting problem. I don't have a general solution, but have the following obvious observations, that might shed light on the problem.

Eqn.1 is that of a plane which goes through each axis at +2

Eqn. 2 is a sphere of radius sqrt(6) centered at the origin.

Solutions must therefore lie on a circle (the plane and sphere could intersect at just one point also)

Given a solution, all permutation are also a solution.

cheers

Dan

quote:

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Originally posted by Dan M:
Ken,

... but have the following obvious observations, that might shed light on the problem.

Eqn.1 is that of a plane which goes through each axis at +2

Eqn. 2 is a sphere of radius sqrt(6) centered at the origin.

Solutions must therefore lie on a circle (the plane and sphere could intersect at just one point also)

Given a solution, all permutation are also a solution.

cheers

Dan

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very interesting observations, especially the last one about all permutations being solutions. this is correct and makes good use of the symmetry of the problem...

i am not sure how much mileage you can extract from the coordinate geometry, but i sense if you try a bit harder, you will get the solution.

enjoy

ken

guys, let me know when you want me to publish a solution to that 3 variable non-linear sim equation...

enjoy

ken

ken,

Please don't wait on my account - I messed around with your equations last night, and came up with nothing of any help, so publish away! (Unless of course someone else is ready with the answer...).

cheers,

Dan

quote:

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Originally posted by Dan M:
ken,

Please don't wait on my account - I messed around with your equations last night, and came up with nothing of any help, so publish away! (Unless of course someone else is ready with the answer...).

cheers,

Dan

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last call ...!!!!

enjoy

ken

last call...!!!!

errr ken,

only you and Dan have played this one so far. although i can't see any neat solutions myself, it might be that Paul Ranson or Omer is simply being polite and giving the rest of us nerds a sporting chance......

another day wouldn't hurt ???

cheers

Don

John,

It's actually 9 in 10,

ooopps! well spotted.....

Cheers

Don

last call ...!!!!

Well, I haven't managed anything very exciting either. I tried the obvious things like trying to combine equations (1) and (2):

(1) (x + y + z)^2 = 2^2

(2) x^2 + y^2 + z^2 = 6

but this just leads to
xy + xz + yz = -1 or x = -(yz -1)/(y + z)
John

quote:

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Originally posted by John Channing:
_last call ...!!!!_

Well, I haven't managed anything very exciting either. I tried the obvious things like trying to combine equations (1) and (2):

(1) (x + y + z)^2 = 2^2

(2) x^2 + y^2 + z^2 = 6

but this just leads to
xy + xz + yz = -1 or x = -(yz -1)/(y + z)
John

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john, you are awfully close!!!

enjoy

ken

quote:

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Originally posted by Don Atkinson:
_last call...!!!!_

another day wouldn't hurt ???

Don

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don, i agree, another day wont hurt... so, get solving!!!

enjoy

ken

Expanding on the fact that (in this case) :

x^3 + y^3 + z^3 = (x + y + x)^3

I (eventually) got to :

0 = 3((x + y + z)(xy + xz + yz) - 3xyz) + 6xyz

Plugging in John's equivalence gives :

0 = 3((2 * -1) - 3xyz)) + 6xyz

0 = -6 - 9xyz + 6xyz

xyz = -2

Not sure if this is worth anything.

quote:

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Originally posted by Minky:
Expanding on the fact that (in this case) :

x^3 + y^3 + z^3 = (x + y + x)^3

I (eventually) got to :

0 = 3((x + y + z)(xy + xz + yz) - 3xyz) + 6xyz

Plugging in John's equivalence gives :

0 = 3((2 * -1) - 3xyz)) + 6xyz

0 = -6 - 9xyz + 6xyz

xyz = -2

Not sure if this is worth anything.

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combined with johns result, that is worth a lot...!!

enjoy

ken

Ok,
xy + xz + yz = -1 can be expressed as

x(y + z) + yz = -1

Using y + z = 2 - x and yz = -2/x we can express this as a cubic equation purely in terms of x:

x^3 - 2x^2 + 2 - x = 0

which can be expressed as:

(x - 1)(x + 1)(x - 2) = 0

or x = 1, -1, 2
John

yep!! well done john. here's is my solution, which shows a slightly different approach, though the end result is the same.

here is a much "simpler" puzzle -- a bit subtle, but no algebraic manipulations.

Lets gather together a group of 12 naim-ites. Now, lets assume that some of them are friends with each other -- well, a safe assumption, isn't it? Show that there will be 2 people who have the same number of friends in the group (which i suppose equates to a good drinking sesh @ next hifi show ...)

enjoy

ken

don, i forgot to reply to your posting about tennis tie break rules, after i stupidly forgot to take them into account in my last tennish puzzle. i more or less copy/pasted this from some usta website -- i think the rules are the same as for wimbledon as well (someone correct me?)

In any set other than final set, if a score of 6-6 is reached, then a tiebreak should be played. In a tie-break, the first player to win 7 points wins the game/set provided that he/she wins by 2 points. If the score reaches 6-6, then the tie-break shall continue until one player gains a 2 point advantage. The player who served the previous game will receive the first point in the tie-break. He/she will serve the second and third points, receive the fourth and fifth points and continue alternating every two points. The first point is served into the deuce court, the second point into the ad court, and continue alternating every point: ad, deuce, ad, deuce, . . . . Thus, other than the first point, each player serves his/her two serves to the ad court then the deuce court (the opposite of what might seem natural). Players should switch sides after every 6 points, and after the tie-break game. The player who serves the first point of the tie-break should receive in the first game of the following set.

enjoy

ken

Ken

I think "let’s assume that some of them are friends with each other" this is the key statement.

If this is true then;

One person must have at least one friend (A is a friend of B) and it may be safe to assume that this friendship is reciprocated then (B is also a friend of A).

From the above then A and B each have at least one friend, so “two people have the same number of friends is true”.

Roy

well done roy.

what about the more general question:

show that at least 2 people will have the same number of friends. i.e. show there is a possibility of more... hence the drinking sesh..


enjoy

ken

the naim party

ken, i have this 'theory of contradiction' which i describe for two people having the same number of friends at the party, but i'm not sure if it easily extends for the possibility of more than two people......

First, we have to define 'friend'. Lets accept it means mutual friends ie if I am your friend you MUST be my friend. We could probably use 'aquaintance'.

Let the number of people at the party be n.

The max number of friends anyone can have at the party is n-1; the smallest number is 0

Let's start by ASSUMING that no two people have the same number of friends

For each person at the party there is an associated 'friend' number (which must be between n-1 and 0 incl)

Each friend number must be different for the assumption to be valid

There are n different numbers to be shared between n party goers

But this is impossible because.......

Someone (say) 'X' scores 0 while someone else (say) 'Y' scores n-1

So Y has everyone as a friend which MUST include X; so X CANNOT have 0 as a score after all.

So we have shown that the assumption that no two people can have the same number of friends is a contradiction and the assumption must be false.

So there must be at least one pair of people with equal numbers of friends at that party (and this must be true AT EVERY PARTY).

Any good ??

cheers

Don

Tennis - again

Now that I have comprehensive details of the 'tie-break' rule (thanks ken) I feel far more confident in posing the following tennis match teaser trilogy....

Tennis trilogy - Part 1

128 contestants enter a tennis tournament (somewhere in south west London - but I forget the exact place). Why 128 ? well, there were a lot more applicants, but the event organisers could afford to be a bit picky and said that 128 would make life easier for them because it would avoid the need for 'byes' (no - it's ok ken, even I know what a bye is.....).

So, after the first round, there were 64 contestants left and so on down to the two semis and the final, when some bloke called 'Pat' (or was it Sue?) was declared the winner.

Part 1 is the easy bit (ok they might all be easy). How many matches were played in this tournament? and can you provide a general formula for the number of matches played in tournaments like this with 32/64/128/256 contestants?

Cheers

Don

Tennis trilogy - Part 2

100 contestant enter a tennis tournament somewhere in (say) Newcastle.....and the organisers couldn't afford to be picky so had to arrange byes for some of the rounds. There were no byes in the first round or second round because there were an even number of players at the start of each round. However, in round three one player got a bye, ie didn't play a match but was entered into the draw for the next round.

How many matches were played in this tournament?

Cheers

Don