Brain Teaser No 1

Just in case anyone was wondering how I did it, this Java code does the trick.
public static void dice()
{
for( int i=1;i<=6; ++i )
for( int j=1;j<=6; ++j )
for( int k=1;k<=6; ++k )
for( int l=1;l<=6; ++l )
{
StringBuffer numberShowing = new StringBuffer();
numberShowing.append( i );
numberShowing.append( j );
numberShowing.append( k );
numberShowing.append( l );
int t = Integer.parseInt( numberShowing.toString() );
int numberOfDice = i*j*k*l;
int tt = i+j+k+l+14*numberOfDice;
if( t==tt ) System.out.println( numberOfDice + ":" + numberShowing.toString() );
}
}
John

John,

Neat.......both the answer and the method (but I haven't checked your method)

Cheers

Don

Omer,

Care to fill us in on the inductive proof of your teaser?

cheers

Dan

Omer et al,

Maybe I'll wait a couple more days for Don's return from his vacation to post the solution ?

Back (superb hol).

Had assumed this teaser of Omer's would have be sorted by now. ha! ha!

Will have a think over the next day or so but admit it looks difficult. and if Paul/Dan/Steved and others have not yet been succesful I don't hold much hope.......

Suggest Omer starts preparing the difinitive answer......

Cheers

Don

quote:

---

However a separation of the points is the line for the inductive proof.

---

Omer,

Will this work?

Find a line such that there are an equal number of black and white dots on one side of the line. Naturally this means there's an equal number on the other side. Sub-divide on one side so that again you have an equal number of black and white dots on one side. Repeat until you have one of each and remove that pair. Repeat the entire process.

Dan

quote:

---

All you have left to prove is that there is such a line that separates the n whites and n blacks into (k,k) on one side and (n-k,n-k) on the other, k!=0.

---

Ah! I thought I might get away without having to do that . Help anyone?
I see what what you mean about not needing the 'repeat process' - a bit of a 'duh' on my part.
cheers,
Dan

Omer,
I may be nit-picking, but the point at B is on the line, not on one side or the other. You will therefore never have a situation where you have an equal number of blacks and whites on either side of it. This is fairly obvious from your diagram.
John

Omer,

Nice proof! So what branch of mathematics does this puzzle come from? Topology? Graph theory?

cheers,

Dan

Omer,
Just a quick observation. Any number, e.g. 5, will stabilize, but will be cyclic:
2,2,1
1,1,3
2,3
2,2,1
The sum of rising series has a cycle of length 1.
John

paul m,

too right, your chances are doubled if you always 'change your mind'.

MDs and CEOs hate that one.....

Cheers

Don

A new, simple, relevant, teaser

In a survey of 100 people, 24 said they drank both larger and beer. Twice as many drank beer as drank larger and 7 were tea-total.

How many pure beer drinkers were there? (ie ones who only drank beer)

Cheers (hick!)

Don

'larger'?

I have an answer, but you're not going to like it, plus four...

Paul

'larger'!!!! oooops. must have had two beers too many tonight....

Cheers

Don

I have an answer, but you're not going to like it, plus four...

I feel someone is about to point out an error in the question.....

Cheers

Don

Don,
46?
Steve D

quote:

---

I feel someone is about to point out an error in the question.....

---

You should have stopped asking at 94 people...

Paul

Don,

I always thought lager was a subset of beer. Assumimg you mean bitter (or any non-lager) and lager, then I get 54 who only drank bitter. Here's my "logic":

a= only lager, b= only bitter

a + 24 + b = 100 - 7
24 + b = 2(a + 24)

3a = 93 - 48 = 45
a = 15 => b = 93 - 24 -15 = 54

cheers

Dan

a+b=100-7-24
a+b=69
b=2a
3a=69
a=23
b=46

Paul

I feel someone is about to point out an error in the question.....

I need to clarify the question.

24 people drink both beer and lager (we'll call these tasteless morons)

2x drink beer and x drink lager

Both the x and the 2x include 24 tasteless morons.......

Hope this helps

Cheers

Don

and apologies for any ambiguity

ask a harder one next time!

I think the 24 morons is just to throw you all off.

a = lager drinkers
b = beer drinkers.

We know:
100 = a + b + 7
=> 93 = a + b

we also know b = 2a so =>
93 = a + 2a
93 = 3a
31 = a

therefore number of lager drinkers is 31 and the number of beer drinkers is 62 and the number of morons is still 24.

Two-sheds,

I think you're double counting the "tastless morons"

Paul,

We agree on a+b=100-7-24, but the second line
is where we differ.

Don,

So which is it?

[ b=2a ] or [ 24 + b = 2(a + 24) ] or ???

To my mind the morons plus beer only drinkers must be equal to twice morons plus lager only drinkers- no?

cheers,

Dan

ok then dan, I'll try not to be too hasty in my assumption of greatness. However here is my revised work.

a = lager drinkers only
b = beer drinkers only
c = morons

1) 100 = 7 + a + b + c
2) 93 = a + b + c

now we also know that 2x drink beer and x drink lager so =>

3) b + c = 2(a + c)
4) b + c = 2a + 2c
5) b = 2a + c

substituting equation 5 into equation 2 gives us

6) 93 = a + (2a + c) + c
7) 93 = 3a + 2c

we know c = 24 so =>

8) 93 = 3a + 48
9) 45 = 3a
10) a = 15

back to equation number 2 and stick in the numbers we know:

11) 93 = 15 + b + 24
12) b = 54

number of pure beer drinkers = 54
number of pure lager drinkers = 15
number of morons = 24


er so i'll just agree with you then dan...

From Don's clarification the answer is clearly,

(100-7+24)(2/3)-24

Paul

clearly

Obviously no further clarification is required thanks to Dan, Twosheds and Paul. Apologies to Steved if the original wording was unclear.

Cheers

Don