I think I've got the full "spark" bit also now.
Bright Sparks
- Thread starter dizz
- Start date
I think I've got the full "spark" bit also now.
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HA - yes, didn't check my answer - perhaps someone can see the schoolboy error?
Dizz,
Yes your Brick answer is correct , but remember it is better spoken than written.
The sheep one I solved as a young kid before I knew simultaneous equations.
One I did know it was easy to use them.
(Whoops you made an error, but then to make things worse you did not check back to see if 8 & 10 fitted in - very naughty - go to BAS and eat some humble pie, then give the correct answer, it is not difficult).
The coin ones with two weighing is dead easy, it is the only one weighing that is bugging me.
No doubt a "Kickself" question as those MENSA guys calls it - once you know the answer you kick yourself.
Another one best done verbal only "Which is greater, six dozen dozen or half a dozen dozen?" It do not really work when written because it makes it far too easy, but spoken is a good un.
Yes your Brick answer is correct , but remember it is better spoken than written.
The sheep one I solved as a young kid before I knew simultaneous equations.
One I did know it was easy to use them.
(Whoops you made an error, but then to make things worse you did not check back to see if 8 & 10 fitted in - very naughty - go to BAS and eat some humble pie, then give the correct answer, it is not difficult).
The coin ones with two weighing is dead easy, it is the only one weighing that is bugging me.
No doubt a "Kickself" question as those MENSA guys calls it - once you know the answer you kick yourself.
Another one best done verbal only "Which is greater, six dozen dozen or half a dozen dozen?" It do not really work when written because it makes it far too easy, but spoken is a good un.
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 piles of coins, with each pile just containing 4 coins and hope to answer it with just two weighings?
The supplementary question was aimed at those that could answer the first and it is a corollary to it, as it's a little more tricky to answer.
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Yes, with all these IQ type tests. the really smart people don't need to work it out, they 'see' the answer immediately after reading the question. Well that is if you define being 'smart' that way, which is debatable in itself.
Likewise, with the original question, the only person on this thread who demonstrated real intelligence was EFLImpudence.
//www.diynot.com/forums/electrics/bright-sparks.330336/page-2#2452302
He 'saw' the answer without any need to get a pencil and paper out.
Dizz,
I will answer the question (although you have "chosen" not to!).
I assume the "accurate scales" are balancing scales.
5 piles,
Take two piles in one pan and two in the other.
If both balance then the pile not on scales is the odd man out and you`ve solved it with just one weighing.
However, if both piles do not balance you have two plies in the heavier pan and you need to balance these against each other to find the heaviest.
That makes two weighings to be certain to find the heaviest pile.
1 in 5 chances you succeed in once.
4 in 5 chances you need twice.
Now then how many sheep?
I will answer the question (although you have "chosen" not to!).
I assume the "accurate scales" are balancing scales.
5 piles,
Take two piles in one pan and two in the other.
If both balance then the pile not on scales is the odd man out and you`ve solved it with just one weighing.
However, if both piles do not balance you have two plies in the heavier pan and you need to balance these against each other to find the heaviest.
That makes two weighings to be certain to find the heaviest pile.
1 in 5 chances you succeed in once.
4 in 5 chances you need twice.
Now then how many sheep?
That certainly is not one weighing!! One weighing mean that you put a coin, or coins, on the scales and observe the (one) displayed answer - and no more than that!
Kind Regards, John.
'Never assume' - any my apologies for not being clearer. NO, I didn't mean 'balancing scales', I meant a measururing device which displays the actual weight of whatever is being weighed.
... but, as you say, if they were balancing scales, you might (4 in 5 chance) need two 'balancings', but could would have a 20% change of doing it with one weighing.
Kind Regards, John.
I can't think of any other meaning, but just to confirm - by an accurate set of weighing scales you do mean something which accurately indicates the weight of something placed on it, or a balance with an accurate set of weights?
Coins II
You have 12 identical looking coins, one of which is a counterfeit, and is either heavier or lighter than the real ones - you do not know which.
You have a simple balance.
Using the balance only 3 times find the fake coin and say whether it is heavy or light.
(BTW - this is easily googled for those who want to cheat).
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, then the gaining farmer will have twice as many as the losing farmer]
The second part of the question implies :
X - Y = 4 ....
[If one farmer gets 2, the other loses 2, making a difference of 4]
From B, X = Y+4 - substitute for X from into [A], then [A] becomes :
Y + 4 + 2 = 2Y - 4
Re-arrange to get Y = 10, and sub for Y back into
so X = 14.
Solution - one farmer has 10, and one farmer has 14.
_____________________
The error was I said X - Y = 2 when it should have been X - Y = 4
Did you REALLY think I couldn't do it?
_____________________
BUT if you read my post regarding the supplementary question to John's I think you'll find you haven;t answered MY question!
Did you realise that I was talking about a weighing device which gives a weight measurement, not just a 'balancing scale', as others have assumed?
Yep, it's certainly one of those!
That's rather different, the issue being that it's very ambiguous (and illustrates why people use mathematical notation!).
Kind Regards, John.
Although ebee made a different interpretation, yes, I can confirm that your understanding is correct.
Kind Regards, John.
Good point - although I suspect that some might work out how to do it from being told that 'minimum'.
That's almost a non-question - since (given only one condition, which doesn't impose a maximum X or miniumum N, per se), my method will always give you the answer with ONE weighing.
However, that makes me think of another way in which I could have asked my question. I could have asked what was the one (and only) condition regarding the number of stacks and the number of coins per stack that would enable one to get an answer with one weighing.
Kind Regards, John.
Well, you've said it, so we'll see - but I think that might possibly 'spoil it' for some people.
Kind Regards, John.
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