probably not right but...

12⁵

I exhausted on 4-long duodecimal numbers. I get 1344=2⁶*3*7, not a power of 12.

I'm very intrigued by this problem and I have no clue as to how to attack it. I'll be working on it and checking back frequently. I've tried to find info about such things on the web with no luck.

If we consider the marbles in a line there are 12C4 combinations = 495.

How many of these are valid? Begin by placing the 4 red marbles spaced out in a line. It is necessary that at least 1 blue marble separates them all. So take 4 blue marbles and put 1 in between each red one (and one at the end which would be in between the first and last). You now have 4 blue marbles left which you can place randomly somewhere between the red ones. The number of combinations with repetition (according to wiki) is in this case: (4+4-1)!/4!(4-1)! = 35. The probability that of no to adjacent red marbles is 35/495 = 7/99.

You get 35 when the chain ends with a red, another 35 if it begins with one, and another 35 if neither the beginning or the end has a red. So, the answer is 105/495=7/33.

The minimum value of X must be

2³*3⁴*223²*251³

in order to make 2008*X=2³*251*X a perfect

square and 2007*X=3²*223*X a perfect cube.

Now, take each factor mod 25 then multiply while taking

partial products mod 25 until you get that X=17 (mod 25).

No, it is easy enough to exhaust over all possible arithmetic sequences with a program. The longest such sequence for primes less than 20000 is the 10-long arithmetic sequence 199,409,619,829,1039,1249,1459,1669,1879,2089.

So, proof here is by exhaustion.

Superprismatic, your explanation is thorough and easy to follow

If Fairy is perfectly greedy/logical, he/she would see that Goblin will vote for any proposal that will net at least 1 piece. Therefore, Fairy only has to offer Goblin that one piece and then sweeten the deal for Vampire, giving [N-3, 0, 1, 0, 2].

Thanks, your modification is better than my solution. I just missed it!

Assume N pieces of identically desired candy pieces where N

is large compared to 5. Below can be easily modified it

they are not identically desired.

If Mummy gets to propose, Vampire will vote against it no

matter what it is so that Mummy will not get a majority and

then Vampire, being the only one left will, of course,

vote on Vampire's proposal. Vampire will propose that Vampire

get it all. So if (F,Gh,Go,M,V) is a vector representing

who gets how much candy, if Mummy gets to propose it will be

(0,0,0,0,N).

Therefore, Mummy does not want the proposing to get to him.

Goblin knows this. If the proposing gets to Goblin, then,

he should propose that Vampire gets nothing, Mummy gets 1

piece, and Goblin gets the rest. He knows that Mummy will

maximize Mummy's candy by voting for this proposal of

Goblin's. In this case the vector is (0,0,N-1,1,0)

Ghost knows this as does Mummy and Vampire. So, if the

proposing gets to Ghost, He should offer Vampire 1 to

get his vote (otherwise Goblin proposes and Vampire gets

nothing). He should offer 2 to Mummy to insure that

Mummy would rather chose his offer to Goblin's possible

offer of 1. He then would get a majority and the

candy payoff vector in this case is (0,N-3,0,2,1).

Now, everybody know this. Clearly, Fairy princess need

only sweeten the offer to Mummy and Vampire to get a

majority. So, she should sweeten Ghost's possible offer

by offering Vampire 2 and Mummy 3 and get the rest for

herself. The candy payoff vector would then be (N-5,0,0,3,2).

9

Let N = 1²+2²+3²+...n

Then, the question is to find the smallest n such that root(N/n) is integer

Let X = root(N/n)

N = n(n+1)(2n+1)/6

Then, X = root((n+1)(2n+1)/6)

This means that n must be odd in order for X to be integer

Let n = 2k+1

Then, X = root((k+1)(4k+3)/3)

This means that 4k+3 must be a factor of 3; in turn, 4K must be a factor of 3

Then, let k = 3r

Overall equation now becomes,

X = root((3r+1)(4r+1))

Since 3r+1 can not be a factor of 4r+1, for X to be integer, either r must be zero or both 3r+1 and 4r+1 must be squares

i.e. 3r+1 = a² and 4r+1 = b²

(r = 0 means n = 1 which violates the condition of n>1)

Then, r = b² - a²

Substituting this above, we get

3b² = 4a² - 1

3b² = (2a+1)(2a-1)

Then a = 1 mod(3) = 3x + 1

Substituting this, we get

3b² = (6x+3)(6x+1)

or, b² = (2x+1)(6x+1)

This means that both 2x+1 and 6x+1 must be squares too

The smallest possible value of x then is 4

This means that a = 3x + 1 = 13

and b = 15

and r = 56

Then k = 56*3 = 168

n = 2k + 1 = 2*168 + 1 = 337

So, 337 is the smallest value of n for which X is integer

Being over inquisitive, I decided to find the next value of n when X could be an integer!

Well, turns out, for this, b = 209; a = 181 and r = 10920

Accordingly, k = 3r = 32760

And n = 65521

I think I know this number (65521) from somewhere... can any one help and tell me what is special about this number?

You said:

"Then, X = root((k+1)(4k+3)/3)

This means that 4k+3 must be a factor of 3; in turn, 4K must be a factor of 3"

Why did you rule out that (k+1) might be a multiple of 3?

That answer does give an integer, I can't get excel to cooperate to give me all answers leading up to that and I'm losing my patience with it. Did you guys use a different approach or program?

RMS of 1...337 = 195

Yes, I wrote my own,

337

I wrote the program to exhaust over all numbers from 1234 to 9876543210

Not only is 893025=(8+930+2+5)² the maximum such number, it is the only such number!

My knee-jerk response is

(8+930+2+5)²=893025

I'll have to write a program to see if this is the maximum.

No one trying....

I thought it was pretty straightforward, so I didn't bother. But, since no one has tried yet, I'll give it another look and probably post my answer by tomorrow sometime.

Prof. Templeton has a correct answer and psychic_mind has a good proof that 6 of each can't work. All that's left is to see how many ways you can place 5 queens and 5 knights on the board so that no piece attacks any other. Any takers for this?

If anyone needs a chess board

All I can come up with is 4 of each

You can do better than 4 each!

Consider a variant of the 8 Queens

problem: you must place an equal

number of knights and queens on a

chessboard such that no piece

attacks any other piece. What is

the maximum number of pieces you

can so place on the board, and how

many different ways can you do it?

My answer is 41 but I think that's just because there are tricks to manipulating the 9's that I haven't been able to figure out. Any help?

41 = (((9+9+9)*(9+9+9))+9)/(9+9)

195.. used a C code

No, that's not it.

Assuming that when you say "smallest" you mean on a linear scale then

(9*9*9*9*9*9)*(9-9-9) would yield something close to the correct answer.

That's negative. The object is to find the smallest positive integer which cannot be expressed.

I found this puzzle on the ITA

Software web site. It is one of

ITA's "retired" puzzles. The

web site does not give the answer.

I suppose the puzzle is difficult

enough to require writing a program

to solve it. I thought programming

this was quite a little tricky but

fun as well. I have an answer but

bugs have been known to creep into

my code. I hope you enjoy it.

Here it is:

Combining nine 9s with any number of

the operators +, -, *, /, (, ) , what

is the smallest positive integer that

cannot be expressed?

Hints:

1)The answer isn't zero. You can

express zero like this:

(9 - 9) * (9 + 9 + 9 + 9 + 9 + 9 + 9)

Also, zero isn't a positive integer.

2)The answer isn't one. You can

express one like this:

9 - (9 * 9 - 9)/9 + 9 - 9 + 9 - 9

3)It's not a trick question.

4)Be sure to handle parentheses

correctly.

Notes:

1)You cannot exponentiate.

2)You cannot concatenate (for example,

put two 9s together to make 99).

3)The - operator can be used in either

its binary or unary form.

4)Assume base 10.

Because, it seems to me, you could fill all the squares except for two which share an edge.

As there are several sphinges in the picture, it must be in Vegas. There is only one sphinx in Egypt.

1) Select a door (call it D1) (1/9 chance of being right, 8/9 chance of being wrong).

2) Allow 5 doors to close, leaving 3 unselected, and the 1 that was selected with 1/9 chance of being right.

3) Apply traditional Monty Hall problem to 3 unselected doors which have 8/9 chance of being right:

a) Choose a door (call it D2) (middle door of remaining 3).

b) Another door is closed - NOT the original selected door D1.

c) Switch to final door, which is NOT D1 or D2 (call it D3).

Original Monty Hall problem: of 3 doors, you pick door 1. Door 2 is opened. By switching to door 3 rather than keeping door 1, you have a 2/3 chance of being correct.

I believe that the math here is simply 8/9 (chance of one of the three unselected doors in step 3 being correct) multiplied by 2/3 (chance of D3 being correct over D2).

edit: My error may come from not adjusting the probability of D1 being correct when the final door is closed, and instead leaving it at 1/9...

Your logic seems iron clad to me. I'm really interested in getting to the bottom of this. At the moment I'm in programming mode and it's hard to switch my mind to probabilities quickly. I'll mull it all over and get back to you if I get any ideas. This turned out to be a pretty interesting problem. Oddly enough, Tuckleton computed his (different from your) algorithm's success at precisely 15/27. I hope he (or someone else) jumps in and tries to explain this stuff.

Let the compartments be numbered 1 through 9, in order. You begin by selecting compartment 1.

Now, compartments 2,3,and 4 are closed, or compartments 7,8,and 9 are closed. For this example, let's say 2,3,and 4 are closed.

You don't switch.

Now, compartments 5 and 6 are closed, or compartments 8 and 9 are closed. For this example, let's say 5 and 6 are closed.

At this point, you originally chose compartment 1. You may now switch to 7, 8, or 9. You switch to 8.

Now, compartment 7 is closed, or compartment 9 is closed. For this example, let's say 7 is closed.

You switch to 9, with 16/27 chance of being correct.

Let the compartments be numbered 1 through 9, in order. You begin by selecting compartment 1.

Now, compartments 2,3,and 4 are closed, or compartments 7,8,and 9 are closed. For this example, let's say 7,8,and 9 are closed.

You don't switch.

Now, compartments 2 and 3 are closed, or compartments 5 and 6 are closed. For this example, let's say 2 and 3 are closed.

At this point, you originally chose compartment 1. You may now switch to 4, 5, or 6. You switch to 5.

Now, compartment 4 is closed, or compartment 6 is closed. For this example, let's say 6 is closed.

You switch to 4, with 16/27 chance of being correct.

Let the compartments be numbered 1 through 9, in order. You begin by selecting compartment 1.

Now, compartments 2,3,and 4 are closed, or compartments 7,8,and 9 are closed. For this example, let's say 7,8,and 9 are closed.

You don't switch.

Now, compartments 2 and 3 are closed, or compartments 5 and 6 are closed. For this example, let's say 5 and 6 are closed.

At this point, you originally chose compartment 1. You may now switch to 2, 3, or 4. You switch to 3.

Now, compartment 2 is closed, or compartment 4 is closed. For this example, let's say 4 is closed.

You switch to 2, with 16/27 chance of being correct.

I printed out the selection array at intermediate steps after I changed the program to do your algorithm. The selection array does exactly what your examples do. When I run the program, I get something very close to 55.56%. Oddly enough, it looks like the chance of being correct is 15/27, not 16/27. I suspect you made a small mistake in your calculations.

This approach is stick-switch-switch. By sticking to your original pick after the original offering, other than the compartment chosen originally, there will be 3 remaining compartments left, all next to each other. The next choice is to remove your original compartment selection from consideration, and force it to not be closed by choosing the middle compartment of the 3 remaining. You are now faced with the original Monty Hall problem, except for the 1/9 chance that the first compartment you selected was correct. A compartment next to your second choice will be closed, and you then switch to the 3rd of the 3 contiguous compartments, giving you a 2/3 * 8/9 chance.

If that's not clear, I tried to make it clear in my earlier post using compartment numbers with an example.

Give me a few examples, please. I'm confused about the middle compartment business because I think, at that point, there are 2 things in every compartment. Examples will clear things up.