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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
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bobbym wrote:

Do you have some idea in mind?

I tried my longhand idea...and I was wrong. Too much mental arithmetic involved, and too easy to make a mistake. Also too easy to miss some. It can be done...with some sweat and tears (unless you're wide awake for this sort of thing).

I had another idea in mind, but that is a bit tricky too and involves quite a few additions. I think my original idea with LB, which was quite easy to set up and solves quickly, is my best option so far.

Btw, is my answer correct?

*Last edited by phrontister (2010-04-25 16:33:35)*

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**bobbym****Administrator**- From: Bumpkinland
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Hi phrontister;

You mean about the number of integers that don't contain a 0 and have a digit sum of less than 9?

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**phrontister****Real Member**- From: The Land of Tomorrow
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*Last edited by phrontister (2010-04-25 16:58:14)*

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**bobbym****Administrator**- From: Bumpkinland
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255 is correct! The answer is suggestive.

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**phrontister****Real Member**- From: The Land of Tomorrow
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bobbym wrote:

The answer is suggestive.

Something to do with Octal (the number after 8 is 11) or binary (255 = 11111111), maybe?

"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
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Hi;

Also pretty sure there is a generating function for it.

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**phrontister****Real Member**- From: The Land of Tomorrow
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Sorry...that's too advanced for me. I never did any of that at school (up to 4th-year high is my limit).

*Last edited by phrontister (2010-04-25 17:54:08)*

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**bobbym****Administrator**- From: Bumpkinland
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Sorry, it is probably wrong anyway.

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

Is there any easy way to compute the unit digit of the nth triangular number?

A) Yes there is, and I know it.

B) That is not possible you just have to put the number into (n(n+1)) / 2 and do the arithmetic.

C) Yes, use a computer.

A really tough one, only I and 2 other people in the world ( and one of them is an alien ) can get this one.

A believes that it will take an average of 36 throws of a dice to get a (1,1). He bets B that he is right. They roll a dice all day long and get an average that is close to 40 throws so B wins the money. A says B got incredibly lucky or the dice is loaded. B says A is an idiot because the expected number of throws is 42 not 36. Who is right and why?

Hard:

A continues the argument from the last problem. He is really mad that B got his money. B being a magnanimous sort of guy agrees to give A a chance to get his dough back. B says let's play another game.

We both throw a die at the same time. I will take (1,2) and you can have (1,1) we keep playing until one of us gets the two rolls in a row that we need. A says that is ridiculous, it is an even game. B says let's play anyway. A says okay. Is B hustling A or is it an even game?

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**bobbym****Administrator**- From: Bumpkinland
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HI;

Medium hard:

Three cards are dealt off of a well shuffled ordinary deck of playing cards.

What are the chances that the first card is a 5 and then the second card is a diamond and the third card is a 4?

A says) 1 / 991.

B says) That's wrong it is 1 / 663. Because:

B was going to explain more when...

C said) Forget that gibberish. A is right,it is 1 / 991.

Who is right?

There are 15 players on a team roster. I need to select 10 of them for the team, 3 of whom will be co-captains. How many ways can I select my 10 man team with it's 3 co-captains.

A says) 120, I counted them!

B says) think of the team as c,c,c,p,p,p,p,p,p,p,p,p,p,p,p and your pick is:

c,c,c,p,p,p,p,p,p,p so:

10 ! / ( 3! * 7! ) = 120 so A is right.

C says) Oh boy, both of you are wrong it's:

Who is right!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

What is the coefficient of x^2000 in :

A says 3^1000 + 2

B says 2^2000 + 2

C says 2^1000

D says You can't do it. It is too tedious.

Medium:

3 people draw 3 different random numbers. They notice that when they take 2 of them, call them x and y and plug them into:

it is divisible by thirty. They pick 3 more different random integers, same thing.

A) says everytime you pick 3 random positive integers there will always be 2 so that f(x,y) is divisible by thirty.

B) says it is just a coincidence. True, there are no small numbers where that doesn't hold but it will eventually fail.

C) says B is right. Here is a counterexample (18735625374653746532611112342965142999^3 -1, 872653419864765342638499993333333^18, 4653625436475888888888888111111113773^25 )

Original work please, I already know the 2 methods shown to me. So I will be looking for a new approach.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Tough:

A bag contains 9 red marbles and 15 blue marbles. Marbles are drawn from the bag without replacement. When the last red marble is withdrawn what is the expected number of blue marbles that have also been withdrawn? Original work please I know my solution and the other one.

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**ZHero****Real Member**- Registered: 2008-06-08
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bobbym wrote:

Medium:

3 people draw 3 different random numbers. They notice that when they take 2 of them, call them x and y and plug them into:

it is divisible by thirty. They pick 3 more different random integers, same thing.

A) says everytime you pick 3 random positive integers there will always be 2 so that f(x,y) is divisible by thirty.

B) says it is just a coincidence. True, there are no small numbers where that doesn't hold but it will eventually fail.

C) says B is right. Here is a countereample (18735625374653746532611112342965142999^3 -1, 872653419864765342638499993333333^18, 4653625436475888888888888111111113773^25 )

Original work please, I already know the 2 methods shown to me. So I will be looking for a new approach.

If two or more thoughts intersect, there has to be a point!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Yes, thank you, I have corrected the original problem.

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

bobbym wrote:

Tough:

A bag contains 9 red marbles and 15 blue marbles. Marbles are drawn from the bag without replacement. When the last red marble is withdrawn what is the expected number of blue marbles that have also been withdrawn? Original work please I know my solution and the other one.

Now I've seen the answer, there's probably a far nicer way of doing this. But here's my way:

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

That is correct. Nicer, that would depend on what you mean by nicer. There is a formula.

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**bobbym****Administrator**- From: Bumpkinland
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5 die are thrown at once. What are the odds that just 2 die are the same?

a says) I ran a program and the answer is: 3600 / 7776

b says) You don't need a computer the GF is ( a + b + c + d + e + f )^5, just expand that and you are done. By the way it is 3600 / 7776

c says) I used the binomial distribution and got 25 / 216

d says) I did it like this:

e says) a and b are right but no one can expand that expression without a computer.

f says) Takes less than 5 minutes to expand b's expression, but it is wrong anyway 144 / 216 is right.

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**bobbym****Administrator**- From: Bumpkinland
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It's alphametic time: Can you solve both equations simultaneously? The usual rules, using only 0 to 9 for each letter.

ZZZ + KKK + LLL = RSTU

ZZZ + PPP + QQQ = RSTU

a says) I looked it up and it's too easy to do. So I worked on something else.

b says) How do you look it up?

c says) I am an expert on diophantine equations. I composed 11 generating functions that prove there isn't any solutions. I would show you but I hope to publish it. You can buy my book.

d says) I used a computer, there are nine solutions.

e says) That's cheating, I hate computers. I would rather put it up there with Riemanns hypothesis and say it can't be done then get an answer that way.

f says) Who is this chap, Riemann? What forum is he on?

g says) Well if you wrote the program isn't that showing enough initiative?

e says) Wrong! Euler used a quill pen and so do I.

h says) I'd like to be as smart as Euler, Where do you buy one of those quill pens?

.

.

.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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So easy!

A bridge charges 11.00 for trucks, 2 dollars for limos and 1.50 for passenger cars. In one hour the bridge collected 27542 dollars. The probability of a passenger car is 40%. The probability of the other 2 types is greater than 0. The counter on the bridge is broken, What is the minimum number of vehicles that passed over the bridge?

a says) No solution is possible!

b says) Of course there is a solution and I have it. It is 3996.

c says) That is one solution, but not the minimum and I can prove it.

d says) The OP is clearly insane. This is not easy at all! I do agree with b, and I can prove it.

Whom do you agree with. I suggest you agree with b. If you agree with a , c or d, you are going to have to prove it.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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B) submitted this proof:

Prove that:

Of course everyone knows the Nicholas Oresme proof that the harmonic series is never an integer when p >1. But the question demanded a proof that took advantage of the fact that p is a prime. This is what B) did.

Assume that:

for some integer n.

Now multiply by (p-1)! The right side n * (p-1)! is still an integer.

All the terms of the LHS up to the last are clearly integers, the last term is clearly not. The sum of a bunch of integers and one rational is not an integer. So the left side is not an integer, Now had the first statement been true that both sides were an integer. when I multiplied by an integer both sides would have remained integers. Proved by contradiction!

A says) It's brilliant, I used it on a test and my MIT professor loved it.

C says) Proof is garbage, there is at least one major hole in it.

D says) I taught B, he is an imbecile, but the proof is correct.

E says) B worked for me, and he is incompetent, I agree with C. B has never been even close to right before.

B says) I stand by this proof! No, it's not brilliant but it is adequate.

Who is right?

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,084

Hi Bobby,

ZZZ + KKK + LLL = RSTU

ZZZ + PPP + QQQ = RSTU

I agree with b, f, g, h and i (i says he used a calculator, and he also partly used LB). Btw, who is/was Euler?

*Last edited by phrontister (2010-06-10 01:42:19)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

That is correct,Good work!

Btw, who is/was Euler?

A real giant. He is long gone.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 96,647

A cryptogram:

LD GZH QFC'O UV BLCS OEVC FO AVFTO UV RFKHV.

A says) cryptograms give me a headache.

B says) I got it. It was easy.

C says) I used a computer, then it was easy.

D says) Takes 5 minutes.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 96,647

A toughie:

One bacterium is placed in a petri dish that has an infinite food supply and infinite space. Every fifteen minutes the bacterium either dies or splits in two with probability of 1/2. Now the two bacteria have the same property, each one either dies in 15 minutes or splits in two with probability of 1/2 etc. What is the chance that the colony dies out?

A says) How gross, besides math doesn't work in biology.

B says) Why on earth not? It's a tough problem but...

C says) The colony never dies out and I can prove it.

D says) If we could figure it, we would find that the colony is doomed.

E says) Wow! A is really smart.

Pick one and prove he is right! You are disqualified if you pick A or E.

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**If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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This one is really daffy but easy. Should get hundreds if not thousands of correct answers.

What is the next number in the sequence?

3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, __?

A says) 9

B says) 8

C says) -123

D says) 0

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