[engineer question] coins part two

[martyr]

this was supposed to be the question i asked initially, but this is better - kinda step you through it this way.
(this should be easy now.)


you have ten machines producing coins.
two machines are malfunctioning.
you may use a scale to weigh a mass only once.
determine which machines are malfunctioning.

same details as last time.
(http://www.planetside-universe.com/forums/showthread.php?t=25981)

[Strygun]

maybe I'll start doing grammar questions when you finish your engineering questions.

[Everay]

wouldnt the last answer work still?

[martyr]

nope

[Smaug]

wouldnt the last answer work still?
No, because machines 1 and 4 could be broken, thus being 5 grams off, or machines 2 and 3 could be broken, still yielding 5 grams off.

[Jaged]

Can we fit 1000 coins on this scale?

[martyr]

hehe - you can fit as many coins as you want on this scale.
fewer is better.
there are multiple answers.


edit:
i can hear the wheels turning from both smaug and jaged

[Smaug]

ok instead of increasing the amount of coins taken from each machine by one(1 from 1st, 2 from 2nd, etc). Take the previous number of coins taken from the last machine times 2, starting with 1. So 1 from 1st, 2 from 2nd, 4 from 3rd, 8 from 4th, 16 from 5th, etc. It gets rather clumsy at the end, ending up with 512 coins from the 10th, but it works I think.

EDIT: Better method derived: take 1 from first, 3 from 2nd, 6 from 3rd, 10 from 4th, etc.
So you are adding 1 to the number of coins added each time.

[JetRaiden]

soooo....is this like a riddle?

[EineBeBoP]

Marty, are you trying to get a job, but pawning off the application test on us? :ugh:

[Smaug]

Marty, are you trying to get a job, but pawning off the application test on us? :ugh:

omfg hax, eine is right.

[KarlMarxist]

I is Sad. :doh:

[AztecWarrior]

What you do is call the Party to track down the comrade who built the machine incorrectly. The miscreant will be re-assigned to logging. In Siberia.

[martyr]

ok instead of increasing the amount of coins taken from each machine by one(1 from 1st, 2 from 2nd, etc). Take the previous number of coins taken from the last machine times 2, starting with 1. So 1 from 1st, 2 from 2nd, 4 from 3rd, 8 from 4th, 16 from 5th, etc. It gets rather clumsy at the end, ending up with 512 coins from the 10th, but it works I think.

EDIT: Better method derived: take 1 from first, 3 from 2nd, 6 from 3rd, 10 from 4th, etc.
So you are adding 1 to the number of coins added each time.

no need to edit.
this counting method is usually called bitwise (powers of two) - smaug's counting coins in binary. perfect.

i myself went for powers of ten, so i wouldnt have to do any math- they both work very well.

[Smaug]

What you do is call the Party to track down the comrade who built the machine incorrectly. The miscreant will be re-assigned to logging. In Siberia.

DOWN WITH THE PROVISIONAL GOVERNMENT. All power to the soviet!

Bitwise. I knew there was some term for that :p

[SniperDude]

Little late on this, but there's an easier way to do the first one:

Grab a balance scale, take one coin from eash machine (total of 10). Divide up into groups of 3, with one x-tra. Put two groups on either side of the scale, if they even, replace one of the groups with the third, if still even it's the random. If one of the previous were not even, take the lighter group and put one coin (from that group) on either side of the scale. If those two are even, it's the random, if not, it's the lighter one.

Make sense?

L8

[martyr]

sniperdude - the scale you're using is a triple-beam. see part one.

[SniperDude]

I know, I know. just pointing it out.