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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Yes, that B guy is the only person in the civilized world who didn't see that you just have to square 1 / x^2. What a dummy.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi;

Moderate to difficult:

Six ducks are swimming in a pond. You are a duck hunter. Each duck can be hit with probability d / 15 where d is the number of unhit ducks. What is the average number of shots required to hit every duck. You have 30 shells, you are wondering whether that is enough. You ask the experts:

A says) 30 is plenty.

B says) Get more cause you will run out.

C says) Wow! 30 is exactly right.

D says) The question is too difficult to answer.

Who is right?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi mathsyperson;

That's correct. Can anybody say what the average number of shots will be?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

12 black balls are to be distributed into 5 bowls:

Which do you like?

12 black balls are to be distributed into 5 bowls each of which must have at least one black ball

What do you think?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi;

This one should get thousands of correct replies.

The sum of 11 numbers is 91. The sum of the first 6 is 61. the sum of the last 6 is 52. The 2 smaller lists have 1 number in common, what is it?

A says) No way can this be done.

B says) This is easy and I can prove it.

C says) You can get it but only using a computer.

Hide your answer and support it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi mathsyperson;

That's correct. Good work.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Ok mathsyperson and phrontister;

Both answers are right but they are both way too short.

It doesn't even look like you did any work. If you want full credit your methods must be fuller, more authoritative, beefier, puffier. Let me demonstrate.

See, beefier, puffier. The trick is to make simple problems look difficult.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Hi Bobby,

Granted, your solution is much more impressive-looking than ours and any astute examiner would be bowled over by the obvious amount of thought that went into that construction. It's just a pity you'd only get maximum credit for it, and not more.

*beefier*: fatter.

I didn't know this until now, and I'm digressing badly here, but *beef-brained* means thick-headed.

*puffier*: containing more puff.

Probably related to *pollyspeak*.

*Last edited by phrontister (2010-03-27 14:33:37)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

phrontister wrote:

but beef-brained means thick-headed

True enough, all this talk about beef and brains is making me hungry.

A wonderful poster on another forum who is as brilliant as she is charming gave me an interesting idea for another problem:

You have a list of n elements that are integers. You need to know the maximum element in the list. You are not allowed to compare elements with a <, >. You are not allowed to sort the list. Remember eyeballing the list is a compare. How would you find the maximum value?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

It depends on exactly what you're allowed to do.

One way would be to subtract one from every element in the list, and keep doing so until only one of the elements is positive. (Or if they all start non-positive, add one to each element until one of them becomes positive)

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi mathsyperson;

I see two problems with that idea. If you decrement each element by one and you started with all positive elements you will have to compare the elements ( to 0 ) sooner or later which is not allowed. Also you have the problem of figuring out whether to increment or decrement. What if the list contains both positve and negative integers?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Yeah, I thought that might be considered cheating.

"No officer, I wasn't comparing, I was just checking for minus signs! Honest!"

The other one's not a problem though. You increment the numbers iff they are all non-positive.

If any of them are positive, you decrement instead.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi mathsyperson;

That would be okay if we knew something about the elements of the list, but you don't.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi;

Easy one:

How many integers don't have a zero and sum to at most 8?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

After thinking more about my suggestion, if you were allowed to check for positivity of numbers then it'd be fairly trivial. a < b iff a-b < 0, etc.

This would work though:

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi mathsyperson;

I walked through your code with some small examples and that idea seems okay. I am not an expert on Matlab so... Presumably you have tested it exhaustively already. Good work!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

bobbym wrote:

hey!! these Exactly were the formulas i was looking for in this post!!

i had come across the same set of formulas somewhere in the forum long ago but was not able to find it again....

thx!!

If two or more thoughts intersect with each other, then there has to be a point.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I just put code there because I thought it was a good way to explain my idea.

Hopefully even if you don't know MATLAB you can see what I'm trying to do.

That idea is much quicker, but it uses the abs function which could be considered cheating.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi mathsyperson;

That idea is much quicker, but it uses the abs function which could be considered cheating.

Not really, you only couldn't use < and >.

I created the recursive idea because her two relationships ( max2 and max3 ) just seemed to suggest it. If I were doing it I would use your procedural approach over recursion.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi;

Easy:

What is the remainder of

When it is divided by 9?

A says) 2

B says) 5

C says) 1

Justify your answer and please hide it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

bobbym wrote:

Hi;

Easy:

What is the remainder of

A says) 2

B says) 5

C says) 1

Justify your answer and please hide it.

Remainder, i suppose, is used in context of Division right???

Shouldn't it be...

What is the remainder of

divided by .......???

If two or more thoughts intersect with each other, then there has to be a point.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi ZHero;

Sorry, can't even copy my own problem correctly. I have fixed the original post.

Easy:

What is the remainder of

When it is divided by 9?

A says) 2

B says) 5

C says) 1

Justify your answer and please hide it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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