[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/f...ad.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]

Originally Posted by Everay

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?

[Smaug]

Originally Posted by EineBeBoP

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

omfg hax, eine is right.

[KarlMarxist]

I is Sad.

[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]

Originally Posted by 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.

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]

Originally Posted by 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.

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

Bitwise. I knew there was some term for that