Simple, but new, puzzle

It's new because I saw it for the first time today.

There is a huge basket with a capacity of holding "X" balls in it. Initially, there is 1 ball in the basket. In the first second, 1 ball is dropped in, in 2nd second, 2 balls are dropped, in 3rd second, 4 balls are dropped, in 5th second, 8 balls are dropped and so on. At the end of 1 hour, the basket is totally filled.

Questions:
At what time was the basket half full?
What is "X"?
And you'll have explain your answer when you solve it (IF you can solve it). Not just copy-paste :D

I won't answer it, to leave the fun for others, but a better version i like:
You have an empty basket, and infinite balls, labelled 1, 2, 3, ....
at 1 min to midnight, you put in balls #1 -> #10, and take out ball #1
at 1/2 mins to midnight, you put in balls #11 -> #20, and take out ball #2
...
at 1/n mins to midnight, you put in balls #10n-9 -> #10n, and take out ball #n
...
The question is, how many balls are in the basket at midnight??

It will be full when 2^(3600-1) balls have dropped in and half full when, of course, 2^(3600-2) have dropped in, that is, one second before. The problem is that neither my calculator or Excel can deal with such a huge number. I leave it to any one of you.

Quote:

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Originally Posted by Rassis

It will be full when 2^(3600-1) balls have dropped in and half full when, of course, 2^(3600-2) have dropped in, that is, one second before. The problem is that neither my calculator or Excel can deal with such a huge number. I leave it to any one of you.

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If you're answering my post, then your answer is incorrect (Read carefully ;)) And if you're answering sql_lall's post... no, you're not answering his post.

Half full as in exactly half or about half?

Quote:

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Originally Posted by mendhak

Initially, there is 1 ball in the basket. In the first second, 1 ball is dropped in, in 2nd second, 2 balls are dropped, in 3rd second, 4 balls are dropped, in 5th second, 8 balls are dropped and so on. At the end of 1 hour, the basket is totally filled.

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You mean "...in 4th (and not 5th) second, 8 balls are dropped and so on." or what?

Thanks

the basket is full at 1 second before 1 hour, so at 3599 seconds.
X=2^3600 (this is to big to show here, 1084 numbers! :) )

initialy there is 1 ball in. every second there will come 2^(n-1) ball, where n is the time in seconds.
So after 3 seconds X = 1 + 2^(1-1) + 2^(2-1) + 2^(3-1), which is equel to 2^3

Pieter

The basket will be full when 2^(3600-1) balls have dropped in or when the basket contains 2^(3600) and it will be half full 1 second before.

Quote:

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Originally Posted by mendhak

It's new because I saw it for the first time today.

There is a huge basket with a capacity of holding "X" balls in it. Initially, there is 1 ball in the basket. In the first second, 1 ball is dropped in, in 2nd second, 2 balls are dropped, in 3rd second, 4 balls are dropped, in 5th second, 8 balls are dropped and so on. At the end of 1 hour, the basket is totally filled.

Questions:
At what time was the basket half full?
What is "X"?
And you'll have explain your answer when you solve it (IF you can solve it). Not just copy-paste :D

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it was half full 1 second before the hour (59min 59 secs)
X is twice the half full amount (lol)

Sounds like a maths problem from a school. You should be able to work out a formula for it...

The basket is never half full.
Once you've dropped the first ball in the basket, you are always adding an even number of balls.

So the basket always contains an odd number of balls {2X + 1 is always odd}.


So, for it ever to be half full, you would have to add a half a ball at some point in time, {since half an Odd number 2X + 1 is X + .5, where X is an integral number}

Unfortunately, you never add a half a ball.

So its never half full.

Quote:

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Originally Posted by NotLKH

Once you've dropped the first ball in the basket, you are always adding an even number of balls.

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Not so sure about that, because:

Quote:

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Originally Posted by mendhawk

Initially, there is 1 ball in the basket. In the first second, 1 ball is dropped in..

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So there is 1 ball in before start, and in the first second we drop in a odd number of balls (1), so we have an even number!

True!
I missed that.

Thanks!
:)

Quote:

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Originally Posted by mendhak

There is a huge basket with a capacity of holding "X" balls in it.... At the end of 1 hour, the basket is totally filled.

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Obviously, at the end of 2 hours, its also {still} totally filled. {Unless someone empties the baskets balls}

So, I've got to wonder.

Does it achieve fullness AT 1 hour, or is it full previous to one hour?

:ehh:

It is full before the hour is up. :)

So, what are the two answers?

http://img393.imageshack.us/img393/1...sket0gr.th.jpg

and so on.

The questions were:

Quote:

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At what time was the basket half full?
What is "X"?

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Where "X" is the total amount of balls needed to fill up the basket.

EDIT: Also, that is not the right equation. (he said 1,2,3,5...)

Quote:

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Originally Posted by Rassis

It will be full when 2^(3600-1) balls have dropped in and half full when, of course, 2^(3600-2) have dropped in, that is, one second before. The problem is that neither my calculator or Excel can deal with such a huge number. I leave it to any one of you.

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2^(3600-1) = 2552433257170972795355421325784140960050835481970150808649442294891973512827494604713191933662806762 2716757084167988820637284734582230408909045052749932766788707495951115682206258172447933264711372686 9243217354413135743351144389318304154546207069175311471992648868850195027417444917592162965295852296 2426885816262185121963825871922744368708437402924184243246325369191792030596395114379845144177186866 0619577150091173099876270214544294693125296540362391561781815405091731382744595014587916240757662618 9505283853706649277815026281162349719949884832098778817193037571710828685146938306208309931545081221 9732396790803760993676875297786943911396561852909025373216343011052039908804945497095607273400609127 9360093516928041725970215462458104461051510233627913415147445683018873412838401413163428788558822026 5189769314661015745127323726981634906736987528338271354992415696612804093631675665724156441594074093 3592448274654530894070007615534170710981252060140941521353105173279637023684918140433107864260306762 723928686968885099947443629498724829053087360954921496828744984350458645612423282688

this number was calculated with GHCi.

Wrong. Think RABBITS instead of FROGS! :)

Tr333,

Great! Excuse my ignorance but GHCi is what? Some maths software...?

It is still not the answer. Think about it.

I have been trying to respond Mendhak question so far and I have noticed that there are some of us that take the answers mistakenly as corresponding to Sql lall post. Let us put some order on this and accept the rule that each post should only address one single subject. I suggest that Sql lall starts a new thread with his problem and leave this only for those of us who have been dedicating their time and effort to find the right answer to the problem presented by Mendhak, who was the one to start this thread. Thank you all.

Quote:

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Originally Posted by Rassis

Tr333,

Great! Excuse my ignorance but GHCi is what? Some maths software...?

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GHC is a compiler for the Haskell language. GHCi is an interpreter for GHC.

I'll take a stab at this. I think that since the question said 1,2,3,5 seconds and NOT 4, then could it be that balls are only added to the basket if the number of seconds is a prime number?

There are 504 prime numbers below 3600, (including 1), therefore since the number of balls added each time doubles the amount, the additon that is made before the last one must be when the basket is half full.

This happens at 3583 seconds, or 59min 43sec

and X = 5.2374E+151

Is that anywhere close?!

Quote:

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Originally Posted by Andy_P

I think that since the question said 1,2,3,5 seconds and NOT 4, then could it be that balls are only added to the basket if the number of seconds is a prime number? There are 504 prime numbers below 3600, (including 1), therefore since the number of balls added each time doubles the amount, the additon that is made before the last one must be when the basket is half full. This happens at 3583 seconds, or 59min 43sec and X = 5.2374E+151

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I think Andy P gave a correct answer if Mendhak meant prime numbers indeed. If so, it was not clear at all to me – reason why I posted on the 1st of the current month the question "You mean "...in 4th (and not 5th) second, 8 balls are dropped and so on." or what?", but unfortunately this was not answered and my efforts were diverted. It shouldn´t be so. :(

Quote:

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Originally Posted by Andy_P

I'll take a stab at this. I think that since the question said 1,2,3,5 seconds and NOT 4, then could it be that balls are only added to the basket if the number of seconds is a prime number?

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Since 1 is not truly a prime number, this could be some other pattern.

For Example:

1 2 3 5 8 13 21 34 ... X_i+1 = X_i + X_i-1

Or:

1 2 3 5 7 10 13 17 21 ...
add1 add1,add2 add2,add3 add3,add4 add4,...
:wave:

Quote:

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Since 1 is not truly a prime number

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Yes, I wasn't too sure on that point, so you could be right.

If the sequence 1 2 3 5 8 13 21 34 etc is correct, then it follows the Fibonacci sequence, and the total number balls will be 131072, and the basket will be half full after 1597 seconds, which is 26min 37sec

X = 131072

Half full at 1597s = 26min 37sec

Quote:

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Originally Posted by Something Else

1 2 3 5 8 13 21 34 ... X_i+1 = X_i + X_i-1

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That's the Fibonacci (SP?) sequence, right?

Yes, as mentioned in my previous post! ;)

Quote:

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Originally Posted by Andy_P

Yes, as mentioned in my previous post! ;)

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I really should read all the posts before posting :rolleyes: :)

I was just to proud of myself to recognise it :D

Another possible pattern:

At N Seconds:

[ 3*N⁵-40*N⁴+185*N³-320*N²+292*N ]/120

Balls are dropped into the basket.

So 6 Balls could have also been dropped in on the 4^th Second.

:)

It will be half full at 59minutes 59seconds ?

The question was not clear in the first place and I think it should be reformulated.

Found after 18 iterations of the Fibonacci Sequence.

Quote:

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Answer: Basket is half full at 1597 seconds.

And basket capacity, is the sum of all the previous values... 524289

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Quote:

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0seconds - 0.5 balls dropped
1seconds - 1 balls dropped
1seconds - 2 balls dropped
2seconds - 4 balls dropped
3seconds - 8 balls dropped
5seconds - 16 balls dropped
8seconds - 32 balls dropped
13seconds - 64 balls dropped
21seconds - 128 balls dropped
34seconds - 256 balls dropped
55seconds - 512 balls dropped
89seconds - 1024 balls dropped
144seconds - 2048 balls dropped
233seconds - 4096 balls dropped
377seconds - 8192 balls dropped
610seconds - 16384 balls dropped
987seconds - 32768 balls dropped
1597seconds - 65536 balls dropped <--------- Half Full
2584seconds - 131072 balls dropped <-------FULL (less than 3600 seconds)
4181seconds - 262144 balls dropped

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I don't understand the double entry at 1 second.

Lets see.

0 seconds, basket contains 1 ball
1 second, 1 ball dropped, bsket contains 2 balls
2 seconds, 2 balls dropped, basket contains 4
...

so that would make:

0:1
1:2
2:4

IF we are counting the number of balls in the basket, and not the number of balls dropped.

hmmm.

Entries 0 to the first 1 seem to be off in your table.


:wave:

Have to add the initial one (to the one at 1 second) to get the count right.

Yes, you should have read the post carefully. And I think that the question was formulated well. AndyP has the right answer, because we're working with the Fibonacci sequence here.

Cool! :)

I guess my old maths teacher was right after all: 'Always read the question!'