a simple brain teaser

Degag

Skid X wrote: »

8 out of 10 Cats does Countdown is on Channel 4 at 9pm.

Featuring Rachel Riley

Have sky+'d the repeat, will be good **** viewing material later!

sock puppet

Ficheall wrote: »

You haven't said whether the 9th ball is heavier or lighten than the average!
You just said it's different.

Now, I've to meet someone for a stroll, but I'm sure you'll have worked it out by the time I get back

You can send me bluewolf's number as a prize or something :P

This is probably why I didn't get those jobs.

Glass Prison 1214

5 to the power of (4-2) + 3 = 28

Edit: well i'm clearly way behind

flutterflye

42

Micky Dolenz

Glass Prison 1214 wrote: »

5 to the power of (4-2) + 3 = 28

Edit: well i'm clearly way behind

Welcome to last week,

You're hired

Ficheall

kraggy wrote: »

I'll word it my own way.

1. Split the balls into 3 groups called A, B and C, each group containing 3 balls.

2. Put A and B on either side of the scales -> 1 use of the balance.

Scenario a.

If they weigh the same, then you know the heavy ball is in group C.
Remove A and B from the balance.
Place any 2 balls from C on to the balance -> 2nd use of the balance.
If they balance, the last ball remaining is the heaviest.
If they don't, the heavier one, is, er, the heavier one.

Scenario b.

A is heavier than B, then remove B and do the same with A as with C above.

i.e. pick any 2 balls and you will determine which of the 3 balls in group A is the heaviest.

Yeah, but you don't know whether the "odd ball out" is lighter or heavier.
In Scenario b, for example, it could be that one of the balls in A is heavier than all the others OR one of the balls in B is lighter than all the others. You can't tell with only 2 weighings...

The actual solution is quite lovely, but I won't spoil the fun by posting it.


And to think they said a phd in pure mathematics would never have any practical applications... now where's my prize? :P

davetherave

Would it be 4 weighs?
1 2 3 = A
4 5 6 = B
7 8 9 = C

*Check A & B. (1st Balance)
**If A=B then different ball is in C. Check 7 and 8 against each other. (2nd Balance)
***If 7=8 then 9 = Odd ball.
Check 9 against any number to see if heavier or lighter. (3rd Balance)
***If 7/=/ 8 then leave the heavier one on and check against 9.
If heavy=9 then the one you took off would be the odd ball and it is lighter than the rest. (3 Balance)
If heavy/=/9 then heavy ball is the odd ball and it is heavier than the rest. (3rd Balance)

*If A/=/ B, check A=C (2nd Balance)
**If A=C, then B contains odd ball.
Do what you did for C up above. Check 4&5(3rd check), if = then check 4&6(4th Check) else check heavy&6(4th Check)

**If A/=/C, then A contains odd ball.
Do what you did for C up above. Check 1&2(3rd Check), if = then check 1&3(4th Check) else check heavy&3(4th Check).

Ficheall

Egad, he changed the question to a trivial one so that his answer would be correct, spoiling the beautiful problem he had originally posed. You make me sick! :P
Or are you just conspiring with Fate to keep bluewolf and me apart?

davetherave wrote: »

Would it be 4 weighs?

Look, Micky D and I have both agreed that the answer (to the original question) is 3. Why must you doubt us??

davetherave

The original question didn't state whether the 9th ball was lighter or heavier than the others. That is what I based my answer on. How can you find the odd ball and determine if it is lighter or heavier in 3 checks?

Ficheall

davetherave wrote: »

The original question didn't state whether the 9th ball was lighter or heavier than the others. That is what I based my answer on. How can you find the odd ball and determine if it is lighter or heavier in 3 checks?

https://www.youtube.com/watch?v=yQej8AOjEHc

Ah, I won't spoil it. It's very nice though


Edit: I just realised I'm beginning to sound like Biggins...

Amirani

This should be it in 3 goes.


Put the 9 balls into 3 sets we can call X, Y, Z.

(1)Weigh X v Y.

A
If X=Y, odd ball is in Z. In this case, (2)weigh Z1 and Z2. If equal, odd ball is Z3. You can check if it's heavier or light by (3)weighing against any other ball.

B
If X != Y, odd ball is in either X or Y. Choose the heavier set [let's say it's X] and (2)weigh against Z.
B1
If X=Z, then Y has the odd ball and the ball is lighter. (3)Weigh Y1 and Y2. If equal, odd ball is Y3. If not equal, odd ball is the lighter one.
B2
If X=!Z, then X has the odd ball and the ball is heavier. (3)Weigh X1 and X2. If equal, odd ball is X3. If not equal, odd ball is the heavier one.

Ficheall

Now pretend you have 12 balls, one of which is a different weight

Amirani

Ficheall wrote: »

Now pretend you have 12 balls, one of which is a different weight

I'll have a think about this.

In the meantime, for those that don't know about it; What is the probability that in a group of 23 people, at least 2 of them have the same birthday?

Ten of Swords

NiallSparky wrote: »

I'll have a think about this.

In the meantime, for those that don't know about it; What is the probability that in a group of 23 people, at least 2 of them have the same birthday?

50% (actually very slightly more than 50% depending on rounding)

Amirani

Ten of Swords wrote: »

50% (actually very slightly more than 50% depending on rounding)

Why?

Ten of Swords

23 people in a group means 253 possible pairs.

The probability of a single pair mismatch is 364/365 = 99.72% therefore the probability that all 253 pairs mismatch is (364/365)^253 = 49.95%

Based on this the probability of a match is 100% - 49.95% = 50.05%

kjl

Skid X wrote: »

((2*5)-3)*4 = 28

2^3+4x5 boom!