Brain Teaser No 1

The guys down at Salisbury only ever have to weigh things that weigh a whole number of "NAPS". They don't use lbs or kgs or grms etc, just "NAPS".

They have only one pair of scales and only four weights. Well, we would all rather they spent the budget on sound quality rather than fancy scales and weights, wouldn't we? Oh, BTW, the scales are the old fashioned balancing ones with a pan on each side.

By using these weights, on either side of the balance, they can work out the weight of items weighing 1, 2, 3, 4, 5, ......."NAPS", and they have carefully chosen the four weights so that this sequence continues as high as possible with just these four weights.

With the balance and four weights, the guys at Salisbury can work out the weight of a MAXPACK but nothing heavier.

What is the weight of a Maxpack?

Cheers

Don

DON,

I think the answer is 40.

Steve D

I agree, 40.

The fastest I can go around a roundabout almost never has anything to do with the top speed of the car. I rest my case, however weak it may be.

Matthew

Mathew, Steved

Perhaps you are confusing the NAPS/Maxpack problem with this one that I set on 1st January 2002...how time flies....

Easy start to the new year!

I feel we need an easy start to the new year. Seems like we got stuck on a couple of mind-bogglers during the hols (Ok, I got stuck!).
What is the minimum number of weights needed to weigh up to 40kg in 1kg increments? Using the old 'Scales of Justice' or simple balance type of scales

I'll get back to the mind-bogglers later!

Cheers

Don


Cheers

Don

PS. just to avoid any doubt, 40 is not the right answer to the Maxpack teaser.......bloody iritating, isn't it !

Matthew, for ages I have been spelling your name wrongly. Spelling isn't my strong point. Apologies.

Hope things improve.

Cheers

Don

I'll weigh in with 46.
cheers,
Dan

[This message was edited by d marsh on MONDAY 02 December 2002 at 20:26.]

I believe it is possible to get to 80 NAPS using 4 weights,

cheers

Dan

80 is possible as long as we are weighing objects that are a round number of NAPS.

Matthew

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80 is possible as long as we are weighing objects that are a round number of NAPS.

The first line of the puzzle reads......

How many NAPS in Maxpack ?

The guys down at Salisbury only ever have to weigh things that weigh a whole number of "NAPS".

So we seem to have crept up from 40 through 46 to 80.

Who would like to offer an explanation, or does anybody wish to offer another number ?

No rush.....

Cheers

Don

Well,

As I see it, the resolution of the scale need be just 2 NAPS, since one can bracket any multiple of NAPS (less than 81), and so uniquely determine its weight. To give a concrete example, say an object weighs 7 NAPS, then it will weigh more than 6 and less than 8 NAPS. Therefore, given that weights 1,3,9, and 27 allow counting to 40 at 1 NAPS resolution, then weights 2,6,18, and 54 allow counting to 80 at 2 NAPS resolution.

cheers,

Dan

Same rational as Dan.

Matthew

Matthew, Dan,

Well done.

Yes, by selecting weights of 2, 6, 18 and 54 (ie double the 'usual' 1, 3, 9, 27) you can precicely weigh all the even weights 2, 4, 6, 8......78, 80. in addition if an item weighs 23 NAPS, say, then by tipping, rather than balancing, Naim can show that the item is heavier than 22 but lighter than 24 and so work out that it must be 23. The Maxpack, or heaviest item they weigh is 80 NAPS.

Just of for a quick nap before returning to work.

cheers

Don

All the characters in this trilogy are fictional. Any resemblance to persons living or dead is purely co-incidental. Apologies to anybody bearing any resemblance whatsoever to Murco, Vic, Perry or Troy.

Murco, the artistic dealer, has been up to his old tricks again. This time with a pack of cards. For the purposes of this quiz, we can trust Murco implicitly.

Murco has shuffled the normal deck of 52 cards and then looked at each one in turn without letting Vic see them. As Murco looked at each card, Vic had to guess whether it was black or red. Murco gave no indication at all whether Vic was right or wrong. After 51 cards, and of course 51 corresponding guesses, Vic had guessed 'black' 26 times and 'red' 25 times and Murco announced that Vic had exactly 10 predictions wrong so far.

Which colour (if either) is more likely to be Vic's last card ?

Part ii will follow after we've had a bit of reaction to part i. Part ii involves Perry, who happens to be a high flung CEO from Swindon, Wilts, UK with a passion for whisky, guns and sorting out rif-raf.

Watch this space !!

Cheers

Don

if it takes 4 men a week to walk a month,how many apples are in a minute?

quote:

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Which colour (if either) is more likely to be Vic's last card ?

---

Red. Inevitably.

Paul

A play on words ?

I'll have to take a walk, for a few minutes, munching an apple along the way hoping for inspiration..

Cheers

Don

Paul R, without commenting on your response, because others may be comming to alternative conclusions, here is the second part of the trilogy.....

All the characters in this trilogy are fictional. Any resemblance to persons living or dead is purely co-incidental. Apologies to anybody bearing any resemblance whatsoever to Murco, Vic, Perry or Troy.

Murco repeated the game with Perry. After 51 cards, Perry had also guessed 'black' 26 times and 'red' 25 times and Murco said that Perry had exactly 11 predictions wrong so far.

Which colour (if either) is more likely to be Perry's last card ?

Part iii will follow after a short interval. Part iii involves Troy, who happens to be professor of medieval English at the University of life in middle England. To while away his sleepless nights whilst analysing early English poetry, he drives the streets of Derby looking for inspiration.

Watch this space.

Cheers

Don

Black. Inevitably...

Paul

i agree with Paul Ranson on both answers. If the responndant guessed 26 times red and 26 times black, then the amount of he would guess incorrectly has to be an even number.

You can therefore know for certain which card the last one is in both instances. No guess work required.

Regards

Belsizepark

i agree with Paul Ranson on both answers. If the respondant guessed 26 times red and 26 times black, then the amount of he would guess incorrectly has to be an even number.

You can therefore know for certain which card the last one is in both instances. No guess work required.

Regards

Belsizepark

Well, the easy part is now over, but you might like toexplain the reasons behind the dead certs and establish the 'rule' about 'red' or 'black' for the benefit of others who might like to join in part iii.

All the characters in this trilogy are fictional. Any resemblance to persons living or dead is purely co-incidental. Apologies to anybody bearing any resemblance whatsoever to Murco, Vic, Perry or Troy.

Murco repeated the game yet again, this time with Troy. After 51 cards, Troy had also guessed 'black' 26 times and 'red' 25 times and Murco said that Troy had got at least 10 predictions wrong so far.

Which colour (if either) is more likely to be Troy's last card ?

Cheers

Don

Now don't all rush.....

Maybe I can't count, but wouldn't 'at least 11 wrong' be more interesting?

Paul

Not sure I like this question.

If he has an even number of incorrect answers we an know for certain that the remaining card is red and if an odd number incorrect the remaining card must be black.

We know that the respondant has at least 10 incorrect and it is theoretically possible he has all 51 incorrect. What hasn't been disclosed is the probability distribution Murco assigns to this.

If the probability distribution (not defined) mean that there is an equal chance of any number from 10 to 51 inclusive of being incorrect then there are 21 even numbers and 21 odd numbers, we can not therefore say what colour has more chance of being the last card.

On any individual card it maybe argued that Troy is right or wrong with a 50% probability (like a coin toss), this means that a bernoulli distribution is appropriate. As there is more than 1 trial then it in effect becomes a binomial problem. However we do know something, there are 26 red cards and 26 black cards. This means that it becomes a permutation problem as opposed to a combination problem. We do not know for certain however if say on the first card Troy has a 50% probablility of being accurate as the test may be some sort of psychic test and Troys actual probability of being right on any given card may be greater for example than 50%

The chances of Troy for example having 51 incorrect cards (before the information given of at least 10 incorrect) is much less than the chance of say having 20 incorrect cards as there are no permutations with every card incorrect. The probability is (1-P)**51 of all card incorrect where P is the probability of him accurately guessing a single card. We can not say from the information given if P is 0.5. There should be a substantially higher probability of 20 cards incorrect. (This is actually intuitive)due to the additional permutations. We could calculate all probabilities quite simply . Additional information is confusing in statistics as Bayes showed with his theorem.

Given the statement by Murco, as no probability distribution is assigned I think we need to ignore the information and state there is an equal chance of the last card being red or black. I would guess that this is the answer that is wanted for the question, as opposed to some long equation. In the event P is 0.5 then the actual answer will be intutively very close or equal to an equal chance anyway.

I will detail an information problem in a seperate message about envelope switching. That problem I think highlights the information issue better than other question I have seen.

Regards

Belsizepark

[This message was edited by belsizepark on TUESDAY 10 December 2002 at 10:18.]

Part 1

Tom holds 1 envelope in each of his hands, and tells Harry that he can have either one of them at his choice. Harry picks one and then Tom tells Harry that one envelope holds twice as much money inside it as the other. Before Harry can open the envelope and see what is inside, Tom offers Harry the chance to switch envelopes. The question is should Harry switch and does it matter?

Part 2

Tom holds 1 envelope in each of his hands, and tells Harry that he can have either one of them at his choice. Harry picks one and then Tom tells Harry that one envelope holds twice as much money inside it as the other. Tom tells Harry that inside the envelope Harry has picked is £100.00. Tom then offers Harry the chance to switch envelopes. The question is should Harry switch and does it matter?

Regards

Belsizepark

Paul R

Maybe I can't count, but wouldn't 'at least 11 wrong' be more interesting?

Possibly, but i've already put some effort into the answer of 'at least 10....!!'

However, the answer to 'at least 11....' would be ok by me. You might have to enlarge the margin a bit to squeeze the answer in....

Cheers

Don