Brain Teaser No 1

p = 91/208 * 288 = 126

But no obvious 'bullet in the head' stuff.

Paul

But no obvious 'bullet in the head' stuff.

Well....that depends.....

for example, your equation for p can be reduced mentally by recognising that 13 is a factor in both 91 and 208, while 16 is a factor in both 208 and 288. This makes the arithmetic relatively simple.

The same recognition patterns can be used to establish possible dimensions for the various rectangles on a 'trial and error' basis. eg 91 has 'neat' possible sides of 7 and 13; 208 doesn't have neat sides of 7, 14 etc but does have 13 and 16 for example etc etc.

So I still think it was a 'do it in your head' type question, if somewhat mundane, possibly. Certainly not worth a 'bullet in the head' though!!

Now the cost of the numbers 'one' to 'twelve' doesn't have an elegant solution, so far as I can tell....

Cheers

Don

I don't see '13' in '91'. You had to be really advanced in my primary school to be allowed to do your 13 times tables....

(I never learned my tables, I just knew enough to be able to work them out on demand. And I could obviously add up. So 13 is really bad news for knowing the numbers.)

(Why did tables stop at 12?)

BTW TWO-ONE = TWELVE-ELEVEN = TW-NE
which looks good for those four.

Paul

What does the "bullet in the head" thing mean ? Did I say something dumb ?

quote:

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BTW TWO-ONE = TWELVE-ELEVEN = TW-NE
which looks good for those four.

---

Paul,

I noticed the same thing, neat! This suggests TEN can be ruled out:

7=TEN-TWO-ONE=-W-2*0

-Dan

This looks a bit suspect :

S+E+V+E+N + T+H+R+E+E = T+E+N

or

S+V+E+H+R+E+E = 0

Actually,

Looking at the numbers from ONE (and assuming that none of the letters is free) :

O < 1
N < 1
E < 1
T < 2
W < 2
H < 3
R < 3
F < 4
U < 4
I < 5
V < 5
S < 6
X < 6

etc

It all works except for :

NINE : < 1 + < 5 + < 1 + < 1 = < 8.

and

TEN : < 2 + < 1 + < 1 = < 4.

quote:

---

What does the "bullet in the head" thing mean ? Did I say something dumb ?

---

You had a small typo; the 'bullet in the head' was to do with flashes of inspiration or being good at mental arithmetic.

Good stuff with the numbers, impeccable reasoning.

Paul

Neat solution Minky.

There is enough info to work out possible prices for each of the letters involved in one to twelve........

Cheers

Don

Pascal,

I happened by chance the other day to discover a rather neat link between Pascal's triangle and the Fibonacci series.

Any bright ideas ?

Cheers

Don

A reminder

Pascal's triangle looks like this

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
etc

and the Fibonacci series looks like

1 1 2 3 5 8 13 21 ...etc

a spreadsheet might help.

Cheers

Don

PS looks like my nice neat 'equalateral' Pascal triangle was 'transformed' by infopop into an ugly rightangled triangle.........but you get the drift

Primary School - Again

This one should be dead simple. Its all about your Ps&Qs

Cheers

Don

Don,
P=8,Q=2,R=4,S=3,T=6,U=1,V=9,W=7,X=5

Steve,
Well done,

Cheers

Don

Any thoughts on pascal and fibonacci?

cheers

Don

A disease ridden probability.....an epidemic or what?

The following are facts (well, for the purpose of this teaser.....so they are non-contestable...)

We have a test for a certain disease that will certainly yield a positive result if the patient has the disease. But it also has a 5% probability of a positive result if the disease is absent.

It is known that 1 person in 1,000 has the disease.

The problem is as follows.

A randomly selected member of the population, tests positive. What is the probability that this person has the disease?

Cheers

Don

Still around 1 person in 1,000 I would guess,

cheers,

Dan

Still around 1 person in 1,000 I would guess,

lets see what others have to say before I reveal the real answer...

Cheers

Don

We have a test for a certain disease that will certainly yield a positive result if the patient has the disease. But it also has a 5% probability of a positive result if the disease is absent.

This question reminds me of an example in the book A Mathematician Plays the Market by John Allen Poulos. I'll dig it out as another brain teaser later.

Anyway, we know the person has tested positive and this could be the result of two scenarios:
1) They have the disease
2) It is a false result which occurs for 5% of the people without the disease

Given that 1/1000 people have the disease, if we use these 1000 people as our sample we will get ~51 people who show up positive on the test:
1 person who has the disease
5/100*1000=50 who will give a spurious result. Of the 51 people who give a positive result only 1 has the disease so the answer is 1 in 51.
John

Well perhaps I was a little hasty I suppose there's some value to the test. I'm inclined to agree with John, but arrived at the answer in a different manner --

A - has the disease
B - tests positive

P(A) = 1/1000.

P(B) = (1 + 999/20)/1000

(1/1000 will always test positive + 5% of 999/1000)

P(A&B) = 1/1000.

P(A|B) = P(A&B)/P(B) = 20/1019

Almost but not exactly 1/51 -- can someone explain?

cheers,

Dan

quote:

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Almost but not exactly 1/51 -- can someone explain?

---

from the 1000 people (if they are according to the stats) we will have:
1 person with the disease
999 people who are not infected

so the number of positive tests it would give for the group would be 5% of 999 plus the one positive which is
((999/100)*5) + 1 = 50.95

so the exact number would be 1 in 50.95

so the number of positive tests it would give for the group would be 5% of 999 plus the one positive which is ((999/100)*5) + 1 = 50.95


That's correct, I used (1000/100*5) for the sake of simplicity.
John

Pascal and Fibonacci.

I rather like this,

F(n)=F(n)
F(n+1)=F(n)+F(n-1)
F(n+2)=F(n)+2F(n-1)+F(n-2)
F(n+3)=F(n)+3F(n-1)+3F(n-2)+F(n-3)
F(n+4)=F(n)+4F(n-1)+6F(n-2)+4F(n-3)+F(n-4)

etc.

(F(n) is the nth Fibonacci number, the coefficients are obviously binomial....)

Paul

quote:

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That's correct, I used (1000/100*5) for the sake of simplicity.

---

ya lazy so and so

A pair of yellow balls.....

On page 2 of this thread, Bam introduced the game show or 'open the box or change your mind' teaser. I particularly liked that teaser because I now find that it irritates CEOs, MDs and other Captains of Industry who can't cope with the idea of changing their mind once a decision has been taken!!!

So I was particularly pleased to find the following teaser the other day, that I am sure is somehow related.

A black bag contains one red ball and two yellow balls. One yellow ball is labled 1 and the other is labled 2.

My friend drew two balls, at random and simultaneously from the bag. What is the probability that both balls were yellow?

Of course, this is only the first part of the teaser........

Cheers

Don

There is a one in three probability (if both balls are picked out at the same time).