Search results

Ha - oh yes - there is other (obvious) info available to me that enables me to identify precisely the stack with any number of piles ...

Maybe maybe not - I work on probabilities - if you left the question the way it is, someone would post the method of attack, and then 100% of other people would know how to do it immediately. My stating the above, might enable a percentage of people to get it quicker, but still a significant...

I don't think it is a non-question, as I think people that get your question right may get the supplementary question wrong - and it gives them something to do! 1. You agree that there is a maximum number of piles with two weighings? 2. You agree that there is a minimum number of coins...

John - I'd ask people to confirm that they know they can do it with just 10 coins and that they couldn't do it with 12 stacks of 20 coins with one weighing! That way they don't spoil it for others that are still thinking?

I thought I'd give others a go at correcting my error, but you'd rather me answer it. Let one farmer have X sheep. Let the other have Y sheep. The first part of the question implies : (X+2) = 2(Y-2) .... [A] [you take two sheep away from one farmer and give two sheep to the other...

I know the error and the correct result, so I'll let someone else spot my error and give the correct answer. So what's the maximum number of piles and the minimum number of coins in each pile where just two weighting will work? There must be a maximum and minimum? You couldn't have 1,000...

HA - yes, didn't check my answer - perhaps someone can see the schoolboy error?

I think I've got the full "spark" bit also now.

Yes, two weighing with John's numbers - but now I am asking what is the maximum number of stacks and minimum number of coins in each stack where TWO weighings is still possible? 30 stacks of 40 coins, 60 stacks of 70 coins, ... ? What I'm looking for is the MAX number of stacks and MIN...

Let one farmer have X sheep. Let the other have Y sheep. The first part of the question implies : (X+2) = 2(Y-2) .... [A] [you take two sheep away from one farmer and give two sheep to the other, then the gaining farmer will have twice as many as the losing farmer] The second...

Ok - haven't seen it, but I get it. You might change your post to ask posters to post the "minimum" number of coins that needs to be weighed in order to solve the problem - that way you know with a high probability they know how to do it (and subsequent posters will see the minimum and...

Go out for a couple of drinks and everyone's been very busy. John - I'm not the spark here, everyone else is - your queetion seems to me to be a spark question not a maths question - never did that in my degree, but if you'd like to give me a proper MATHS question, Pure, Statistics, Mechanics...

Thanks John - This post was really directed at BAS - I really wanted to see whether he had the balls to put an answer forward, which he didn't - but I bet there are loads of scraps of paper where he'd had a go ... Unfortuntaley if took another poster to show him the answer before he "knew" how...

What did you do, take a look at which poster I "thanked" and then regurgitated stuff like you always do? The reason I gave you this dumb ass question was because it's something I came across in real life - a length of cable unknown in pre-wired conduit in a property which had been extended in...

A big FAT nothing. Looks like you had a go at posting something, but decided to delete it - after all, if your calculation is WRONG for the length of cable, that could be DANGEROUS. So, just because you are familiar with EIC forms, and know how to complete them that makes you a compentent...

A troll that's been a member for three years? You know what BAS, you're just slightly beginning to give the impression you cannot do this simple question?

So prove yourself BAS - Here is a pretty simple question that I am sure you can answer - so what's the answer?

1.5mm = 12.1 2.5mm = 7.41 That should be all you need - Anyone?

Here's a question to all you bright sparks out there. I've made it up so you won't find the question or answer on google. An unknown length of cable has all conductors with csa of 2.5mm. The circuit is extended with an unknown length of cable with Live conductors of 2.5mm and CPC of 1.5mm...

Grow up BAS - you really think it's dangerous if that I don't know what "In" means for the form if I can correctly calculate voltage drop, CCC of cable depending upon installation method, and then selection of correct MCB based upon PFC and demand of appliance on circuit? GAGYSMW.