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 (2010-04-26 14:33:35)

Last edited by phrontister (2010-04-26 14:58:14)

bobbym wrote:

The answer is suggestive.

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

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 (2010-04-26 15:54:08)

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?

Hi;

What is the coefficient of x^2000 in :

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.

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:

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:

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.

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 23:42:19)

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.

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