Puzzle

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Fran is off on a trek across the Sahara desert to see Ricky Martin in concert. It is a six-day hike from the airport to the closest oasis where he is playing. One person is only physically capable of carrying enough supplies (food, water, Ricky CD's) for four days. Obviously, she cannot make the entire trip alone. How many assistants will she need to make it to the oasis while making sure that the assistants have enough supplies to make it back to the airport? (They don't want to see the concert.)


a/ 4

b/ 3

c/ 2

d/ 1

e/ how is this related to podiatry?

 

f/ why would she want to see Ricky Martin anyway?

 

smart bottoms, it's obvious but just lets pretend..

 

if the assistants leave her at the concert then she'll die on the way back as she can only carry 4 days worth of provisions for the 6 day return journey...... unless Ricky decides to bring her back in style :drinks

 

ok ok so anybody with such poor taste deserves to be left in the middle of the desert...
i give...........i got it wrong and i didn't have any disturbing distracting thoughts that you do mr/mrs/ms T kemp!

 

2 assistants















don't shoot me if i'm wrong

 

2 assistants

or 3 if the 1st assistant does not want to go back out into the desert.

 

Tkemp, did you read Peters post, no muckin' around, straight to the answer "2 assistants", just like that......now next time.....and then I would get to reply
Correct! thanks for playing Peter ........................................(but I'd say Tkemp not Peter, that is unless you are Peter?)

 

LOL!!!
but she is still in the desert after the assistants have returned

sorry... couldnt resist
and it's Tracy ....... though Ms Kemp is also acceptable, along with Almighty Supreme One of Peanut MnM's, Ruler and Protector of the Coffee Jar, etc, etc
:drinks

 

Since everyone enjoyed that last puzzle, here's another, an old one:

You are in a house which has a room upstairs which has 3 light bulbs. Downstairs there are 3 switches, each corresponding to one of the three light bulbs.
You can turn the switches on and off and leave them in position.
If you're only permitted one trip upstairs, how would you identify which switch corresponds to which light bulb?
Be good Tracy.

 

walk upstairs, leave the door open and shout to your other half to flick the switches

 

i do know the answer...... honest

 

Yes Tracy, that would work, if your 'other half' wasn't a 'half wit', and most men do have their shortcomings so i'm regularly reminded,.......... soooo, qualification is that you're all by yourself, you can't see the upstairs room from downstairs, also, just in case, it doesn't matter what country this house is in, nor the colour of the walls, nor if you were in an inquisitive mood or not, and oh yes there isn't a 'power out' at the time! (who'd want to be an administrator)

 

ok ok......
you turn on two switches and wait 5 -10 mins.
then you turn one off and quickly go upstairs.
The light bulb on is the switch left on.
The light bulb off but warm/hot is the switch you just turned off.
The light bulb off and cold is the third switch.

................. children please do not attempt this without a parent nearby. Hot bulbs can burn!

 

Correct! thanks for playing Tracy........now wasn't that worth it

 

now for a difficult puzzle, (so they say though i still haven't got one correct)

There are 20 imprisoned felons awaiting execution to occur on the next day.
The governor addresses them,

“I am going to give you felons a chance to be set free tomorrow. You will all stand in a row (queue) before the executioner, all facing one direction, with each felon only being able to see the felons in front of him. We will place a hat on your head, either a red or a black one. Of course you will not be able to see the color of your own hat; you will only be able to see the prisoners in front of you with their hats on; you will not be allowed to look back or communicate together in any way (talking, touching.....)

(The prisoner in the back will be able to see the 19 prisoners in front of him
The one in front of him will be able to see 18…)

Starting with the last person in the row, the one who can see everybody in front of him, he will be asked a simple question: WHAT IS THE COLOR OF YOUR HAT?

He will be only allowed to answer “BLACK” or “RED”. If he says anything else you will ALL be executed immediately.

If he guesses the right color of the hat on his head he is set free, otherwise he is put to death. And we move on to the one in front of him and ask him the same question and so on…"

The felons can communicate freely during the night. Is there a method to guarantee the freedom of some prisoners tomorrow? How many?

 

It has made my day worth living

 

It's been 'brought to my attention' that apparently Peter always answers, "Two assistants", to any question he's asked, no matter what the subject. This will not do young Peter, the kudos has been withdrawn, how embarrassing for you!

And Tracy, lovely to hear from you.

 

In the evening, the prisoners decide that they will call out the colour of the hat of the prisoner in front of them. This means that prisoner 19 to 1 will survive with a 50% chance of prisoner 20 surviving.

 

Still thinking

 

they need to know the number of each colour of hat first ... then it depends on the first one asked to be able to count the colours in front of him

 

gday Trevor, i couldn't get anywhere with this one so don't think i'm a knowall, had to look at the solution.
you got the first bit in that the prisoners did certainly talk and decide on a strategy.
but calling out the colour of the hat in front means the first 19 to call have a 50% chance whilst only the last gets it 100% correct.

yep, updates are always appreciated, thanks Trev

very close, i think, but the first felon to answer ie the last in line, who happens to be a good bloke, his response is most important for all

 

Prisoner 20 will call out the colour that is ODD.
ie If he can see 7 black and 12 red hats, he will call out BLACK. He has a 50% chance of surviving.
BUT, prisoner 19 will see either (A) 7 black and 11 red or (B) 6 black and 12 red. As he knows that including his, there are an odd number of black hats, then he will call out RED if (A) or BLACK if (B).
Prisoner 18 would see (C) 7 black, 10 red, or (D) 6 black, 11 red.
He knows from prisoner 20 that there was an odd number of Black hats and if prisoner 19 called red, then he would know there are still an odd number of black hats including his, so he would call BLACK if (D) and RED if (C).
If prisoner 19 called black, then he would know there is an even number of black including his own, so he would call BLACK if ()C) and RED if (D).
etc.
ie 19 would survive for sure, with number 20 a 50% chance.
Hope this makes sense.

 

not true? should've realised, Peter, mate, you've got enemies

Trev, more sense than the answer that was published as the official solution.

Trev d'man, respect

 

Thanks Mark, now here is one for you.

2 podiatrists, Alan and Bob hadn't seen each other since their uni days. They met at the National Conference and started chatting.

Alan: Hi Bob, how are things going? Do you have any kids?
Bob: Yes, I have 3.
Alan: How old are they?
Bob (being a man who likes numbers): The product of their ages is 36.
Alan: That doesn't tell me their ages.
Bob: Sorry. The sum of their ages is equal to the number of the hotel next door.
Alan went over to the window and checked out the hotel next door. He then came back.
Alan: That still doesn't tell me their ages.
Bob: Sorry. The oldest one plays the piano.
Alan: I see!! Their ages are .... , .... and ....

How old are the children and explain how you got the answer. ie don't just guess.

 

gday Trev

only permutations available for the first bit are
1,1,36
1,2,18
1,3,12
1,4,9
1,6,6
2,2,9
2,3,6
3,3,4

when you sum them you get

38
21
16
14
13
13
11
10

bob or alan ?? couldnt tell after he was given the 2nd clue therefore had to be one of the 13's

the last clue states 'the older' so of the two 1,6,6 and 2,2,9 it has to be 2,2,9


(flat out at work but being superefficient managed to squeeze it in)

 

the product of their ages must equal 36 and added together their ages must make a certain number.
because one child plays the piano he is older than the others.
So if 36 = 3*3*2*2
then the first child is 9 (3*3)
the other two children are twins and both 2 years old.

the hotel number is 13.

 

Well done Mark and Tkemp. Now back to work.

 

thanks Trev, i might be able to post some more for the weekend.
Gday Tracy, if i had known you were a mildura resident i would have called in on our last trip to adelaide and you could've shouted my good wife and i to a sumptous feast at the local,
mark of canberra

 

Once you've feasted in Mildura, the rest of the country seems drab in comparison

 

Just talking about Mildura and I'm getting adrenaline 'rushes'. I'm with you baby!




Puzzle for this weekend: (a very old one, if memory serves me well, I think)

Twelve balls, one of which weighs slightly different to the other eleven. Not told whether heavier or lighter.
How would you identify this odd ball if you could use an old balance scale only 3 times! (old balance scale consists of two sides of a pivot and can only determine if there is a weight difference between the objects on the two pans and which sides are heavier/lighter)
You can only balance one set of balls against another, so no reference weights and no weight measurements permitted.

mark

 

I just choked on my drink!

 

take the 12 balls in half 6 balls on each side.

take the heavier 6 balls half each side again

take the heavier 3 put 2 balls on the scale if the scale moves pick the heavier ball if not the ball left out is the heavier.

 

But what if those are all normal balls, ie a light ball with 5 heavy balls has been excluded in that process?

This is assuming a heavier ball is the outlier (not a light ball)

I don't have the answer yet though, either:wacko:

 

Number the balls 1 to 12. Weigh 1, 2, 3, and 4 against 5, 6, 7, and 8.
If (1, 2, 3, 4) and (5, 6, 7, 8) balance:
Weigh 9 and 10 against 11 and 8 (we know 8 is not the odd ball).
If (9, 10) and (11, 8) balance: then 12 is the odd one.

Weigh 12 against any other to find out if it is heavy or light.

If (9, 10) and (11, 8) do not balance: suppose 11 and 8 are heavier,
than 9 and 10; then either 11 is heavy, or 9 is light, or 10 is light.

Weigh 9 against 10; if they balance, 11 is heavy; if they do not,
the lighter of 9 and 10 is the odd ball.

(Similar argument if 11 and 8 are lighter than 9 and 10).

If (1, 2, 3, 4) and (5, 6, 7, 8) do not balance:
Suppose 5, 6, 7, and 8 are heavier than 1, 2, 3, & 4. Then: one of
(1, 2, 3, or 4) is light, or else one of (5, 6, 7, or 8) is heavy.
Weigh 1, 2, and 5 against 3, 6, and 9.
If they balance: then either 7 is heavy, or 8 is heavy, or 4 is light.
Weigh 7 against 8; if they balance, 4 is the odd ball, otherwise the
heavier of 7 and 8 is the odd ball.

If (1, 2, 5) and (3, 6, 9) do not balance: suppose 1, 2, and 5 are lighter
than 3, 6, and 9; then either 6 is heavy, or 1 is light, or 2 is light.
Weigh 1 against 2 to find out which one of the three choices is true.
Otherwise, suppose 1, 2, and 5 are heavier than 3, 6, and 9; then either 3
is light, or 5 is heavy.

Weigh 3 against (say) 2 to find out which of the two choices is true.

(Similar argument if 1, 2, and 5 are lighter than 3, 6, and 9)."

Love kisses & cheating x

 

the light ball does throw somewhat of a curve ( I admit i did not read Marks post carefully enough ie were not told if there heavier or lighter )

I´ll get my bat and ball and be off then .

 

Mandy's done a Trev out of 'left field', Michael's taken his 'bat'n'ball', Tracy's in ED, Peter's about to put his foot through the monitor, and I'M GOING TO BED! night world

 

Not a clue what one of those is. Though I'm certain it's accurate! Sleep tight x

 

Three boxes are all labeled incorrectly, and you must get the labels right. The labels on the boxes read as follows:

[box 1] nails
[box 2] screws
[box 3] nails and screws

To gain the information you need to move the labels to the correct boxes, you may remove a single item from one of the boxes. You may not look into the boxes, nor pick them up and shake them, etc.

Can this be done? If so, how? If not, why not?