Math problem 83 or something, Real mathematicians can't count Options

[Raijinili]

Say that when you take pistachios, you take ten per handful, you eat the unshelled ones, take off the shells of the shelled ones, and drop the newly unshelled ones back into the bowl.
If you had a bowl of 100 shelled pistachios, how many grabs would it take to eat them all?

[jcdietz03]

after thinking about this for a bit...

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isn't it 20? On each handful you grab ten pistachios. You need to grab each one twice in order to eat it. You need to make 200 grabs total. You make 10 grabs per handful, so you need 20 handfuls.

After thinking about this some more, I think you need 21 handfuls:
Take a simpler case where u have 20 in your bowl. Theoretically, you'd need 4 grabs.
After round 1 you have 10 shelled and 10 unshelled.
Let's say for round 2 you picked 5 shelled and 5 unshelled. After round 2 you have 5 shelled and 10 unshelled.
Let's say for round 3 you picked 2 shelled and 8 unshelled. After round 3 you have 3 shelled and 4 unshelled.
You pick them all up for round 4, but at the end of the round you have 3 unshelled still. You need to make a fifth grab in order to eat them.

This post has been edited by jcdietz03: Feb 2 2008, 11:05 PM

[regruBgniK]

Who eats pistachios without the shell?

[Raijinili]

An example isn't a proof.

[jcdietz03]

An example isn't a proof.

It's been too long since I did proofs.
That's the answer, amirite?

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You need to pick up each one twice in order to eat it. There are 100, so you need 200 grabs.
As long as there are >10 pistachios in the bowl, you'll be able to grab 10, and by doing so, you reduce the "number of operations to go" by 10.

After 19 grabs, there will be <10 pistachios in the bowl. I don't know how I know this. It might be b/c it isn't true.

Here's a try at the proof for that though:
On average, every 2 operations, 1 pistachio disappears. This means that, on average, after 190 operations (19 grabs), there will be 5 pistachios left in the bowl.
100 to start with (minus) (190 operations * 1 pistachio / 2 operations) = 5 pistachios left after 190 ops.

On the 20th grab, you cannot grab 10 pistachios, b/c there are less than 10 for you to grab. Because you pick up less than 10, you can only complete the number of operations equal to the number of pistachios you picked up. Also, all the pistachios in the bowl at this point will be unshelled. On the 21st grab, all the remaining pistachios will be unshelled (because any unshelled ones you got on the 20th grab will now be unshelled), thus completing the task.

[Raijinili]

It's not true that there will be less than 10 after 19.

I also think you shouldn't look at the averages.

[jcdietz03]

It is true. I meant to say "less than or equal to"

The worst case scenario (for most pistachios left after 19 grabs) is if you picked up shelled ones whenever they are available. This will reduce the number in the bowl the slowest.

So after 10 grabs you would have 100 unshelled. After 19 grabs, you would have 10 unshelled left. Any other combination will reduce the number in the bowl faster. There are other permutation that will result in 10 unshelled left after 19 grabs, but there are no permutations that result in more than 10 pistachios after 19 grabs. Or if there are, name one.

If you mixed it up a bit more (left shelled ones in the bowl when they are available) you would have even less than 10 left after the 19th grab, but some would be shelled.

This post has been edited by jcdietz03: Feb 3 2008, 09:48 PM

[Raijinili]

It is true. I meant to say "less than or equal to"

Then what you said was, in fact, not true.

Or if there are, name one.

I'm the one asking the question here. It's up to you to prove that there are none.

[jcdietz03]

This is your claim:
It's not true that there will be less than 10 after 19.

The < vs <= is a technicality. Don't argue technicalities, just provide the correction.

If I change my statement to <=, then it's right. Anyway, I think I did prove it, just follow the logic. The worst case scenario (for maximum pistachios left) is provided. Any other scenario is either the same or better.

I guess I still need to prove that last part though:

If you mixed it up a bit more (left shelled ones in the bowl when they are available) you would have even less than 10 left after the 19th grab, but some would be shelled.

The number of operations left in the bowl is calculated by [# of unshelled]+2*[# of shelled]. Each time you make a grab, the number of operations goes down by 10. After 19 grabs, you will have 10 operations remaining. If there are 10 unshelled in the bowl, you can complete the task on the next grab. If there is 1 shelled in the bowl then there must be 8 unshelled because there are 10 ops left in the bowl, and each shelled represents 2 ops. Meaning nine total are left in the bowl.

Therefore, the quoted statement is proven.

[Raij|Away]

That's right.

The < vs <= is a technicality. Don't argue technicalities, just provide the correction.

The tester doesn't need to tell you exactly what you got wrong. A simple "That's wrong" will suffice. Be grateful that you got more.

[DustyHaru]

Is that shelled, the past participle of 'to shell' (ie, having been shelled, and therefore having no shell) or shell-éd, as in possessing the shell?

(My friend from school asked, that's his typing^)

-Dusty

[Raijinili]

Shelled as in opposed to unshelled. I also specifically stated that you take the shells off the shelled ones.

[Wally]

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No one eats pistachios like that. The question is moot.

[Raijinili]

you moot

[DustyHaru]

BORED

Is it least number of grabs or most number of grabs?