Yes, it made sense. I'm also tired. This will have to wait for 18 hours or so for my attempt to run this thru the program.

But I will.

I fixed the bug and ran with your strategy 100,000,000 times and got 55.561676%. A very good agreement with your analysis, Tuckleton.

Super, any chance you could run my method outlined in post dated 09 October 2009 - 10:18 PM through your program?

59.3%

I'd be glad to try. I'm not exactly sure of your algorithm, though. Please explain in more detail.

sqrt(((sqrt(9))!)! / 20)

FOUL! sqrt uses 2 (much like fortieth root uses 40).

Well that's discouraging. I probably did the probabilities wrong somewhere along the line. Perhaps you could enter my method into your simulator and let me know what it comes out as to help me locate my error.

So pick any door, then, once doors are opened, follow the line of opened doors from the picked door and switch to the first unopened door. Do this again after doors are opened a second time. After the last door is opened, switch to the door that could have been opened but was not. For example, you pick door number 1. The 3 doors on it's right (234) are opened so you go right from 1 to the first unopened door, which is 5. Then 2 doors on the left (19) of 5 are opened so you go left to the first unopened door, which is 8. Finally, the door on it's right (5) is opened so you pick the door on it's left, door 7, and cross your fingers. Let me know if that made sense. I'm kind of tired

Yes, it made sense. I'm also tired. This will have to wait for 18 hours or so for my attempt to run this thru the program.

But I will.

What is the smallest positive integer that

(xthroot(9²⁰))! can be for positive integer x?

(fortiethroot(9²⁰))!

I wrote a simulator to try all ways to stay/switch, each way being tried 10,000,000 times. The best I could do was win 38.4% of the time. Of course, I could have a bug even though the program is pretty simple.

Sorry, posted to the wrong problem!

Dr. Brainard is inventing again. He invented an lazy susan with divided areas that can collapse and have the remaining areas expand across the entire area. Imagine a series of Japanese hand fans (an accordian folded piece of paper that looks like a stick but can stretch out to make a full circle) all connected at the center. He is tickled pink, just imagine one day you can have 2 compartments: 1 for chips and the other for dip and then the next day you can have wings divided into spicy, cajun, hot, bbq, and honey mustard. But no one is buying them so he invented a game to go with it.

The lazy susan starts with 9 compartments and a slip of paper goes in each one. As with all good games, all the slips but one say, "sorry, please try again" and the last one says, "WINNER". As a contestant, you pick a compartment and the host "closes" 3 adjacent compartments to one side or the other of the one you picked, saying/showing you that those 3 were not the winning compartments. Now you have a chance to pick a new compartment or stay after which the host will close 2 adjacent compartments to one side or the other the one you now have picked. This again repeats with the host closing 1 more compartment, leaving 3 compartments left. You have 1 last chance to change your pick.

So to help imagine them numbered 1,2,3,4,5,6,7,8,9 and if you pick 1 the host may close numbers 2,3,4 or 7,8,9.

so you end up with 1,5,6,7,8,9 or 1,2,3,4,5,6 and if you stay with 1 the host may close numbers 5,6 or 8,9 in the first case or 2,3, or 5,6 in the second case.

etc...

The 1st part is, "what is the best strategy? and what is the resulting win probability?"

People are still not buying them so Dr. Brainard decides to try and sell them to casinos. Now being for a casino, he had to come up with payments so being tired he came up with something easy: you pay $1 to play and win $8 if you pick the right answer, but you have to pay another $1 to have the host close compartments and make a new selection. So you pay end up paying $4 if you want to go to the end, but you can stop at any time. So does this change your strategy and what is your best ROI?

I have a question whose answer will affect the probability.

If the host can choose either of the two groups of compartments

on either side of your current choice, does he do so randomly?

Suppose the winning number is 9 but you chose 4. The host

closes 1,2,3. Now, you change your guess to 6. The host may

wish to close 7,8 rather than 4,5 to make you think that the

4 that you just abandoned may actually be the winner. This

may entice you to go back to 4 later. If he's allowed this

kind of shenanigans rather than pick randomly, the

probabilities will be affected.

is there a limit to the integers? can they be negative?

There is no upper limit to the integers. The OP does specify positive integers, so the lower limit is 1.

Consider the following two-player

game: Each player secretly chooses

a positive integer. The integers

are revealed simultaneously. The

object of the game is to choose an

integer which is either exactly one

larger than your opponent's integer

(in which case, your opponent pays

you $2) or which is at least two

smaller than your opponent's integer

(in which case, your opponent pays

you $1). If you both choose the

same number, the game is a draw and

no money changes hands. So, for

example, if A picks 6 and B picks 5,

B gives A $2; if A picks 6 and B

picks 2, A gives B $1; if A picks 6

and B picks 7, A gives B $2; if A

picks 6 and B picks 8, B gives A $1.

What is the optimal stratagy for

playing this game? Why?

sp - I guess I assumed that KS was using a generic definition of "magic square" to mean a square with all rows, cols and diags to result in the same answer and then defining the +,-,+ rule instead of a simple +,+,+ rule.

Oh, now I know how you interpreted the OP. I read it as the +,+,+ rule has to work as well as the +,-,+ one. I guess only KS knows for sure what he meant! Please enlighten us, KS.

0. I agree with tbrophy. The squares in posts by psyclist and Doctor Moshe are not magic squares.

Determine the total number of positive decimal integers of the form 10^m - 10ⁿ that are divisible by 1001, whenever m and n are positive integers such that m < n < 100.

What is the total number of positive decimal integers of the form 10^m - 10ⁿ that are divisible by 239, whenever m and n are positive integers such that m < n < 100?

As stated, the answer is 0. However, if instead, we use n<m<100 the answer is 768 (the number of ways m-n is a multiple of 6) to the first question, and 651 (the number of ways m-n is a multiple of 7) to the second question.

i checked up to 100,000, as far as i can tell there is no other solution.

Did you check to see what's the closest you got to a solution, i.e., what is the closest the right and left hand sides got to each other? It would be interesting to know! If you got as close as 1 or 2 or so, you may not have gone out far enough.

Since the function F(X,Y)=Y⁴+(Y+1)⁴-X²-(X+1)²

is a differentiable function of 2 variables, an extremum point can be found in the usual

way by taking partials, setting them to 0, and solving. One must also check boundary

points which in this case is (X,Y)=(0,0). Doing this, we find that F(X,Y) is minimal

at only the boundary point. There are other candidates from doing the partial derivatives,

but they are not non-negative integers. So, (0,0) is the only solution.

Well, I forgot to actually check the extremum candidate from the partials. It would have

to make F(X,Y) greater than 0 for my remarks to make sense, i.e. for the solution over

the reals to be no better than the boundary candidate. As it is, the candidate from the

partials makes F(X,Y) take its smallest value of -3/8 at (-1/2,-1/2). So, my previous

post is incorrect.

Since the function F(X,Y)=Y⁴+(Y+1)⁴-X²-(X+1)²

is a differentiable function of 2 variables, an extremum point can be found in the usual

way by taking partials, setting them to 0, and solving. One must also check boundary

points which in this case is (X,Y)=(0,0). Doing this, we find that F(X,Y) is minimal

at only the boundary point. There are other candidates from doing the partial derivatives,

but they are not non-negative integers. So, (0,0) is the only solution.

The difference of two base 10 numbers which are digit permutations of one another is always a multiple of 9.

Note that, for a difference of two powers of 2 to be a multiple of 9, the powers must differ by a multiple

of 6. This is easy to see if you think of the two numbers in base 2 -- 9 is 1001 and the difference of the

powers of 2 is a sequence of 1s surrounded by 0s. But, powers of 2 differing by a multiple of 64 (2⁶)

makes them differ in the number of decimal digits they contain. Therefore, they cannot be permutations of each other.

Is your answer a "hunt and peck" method, SP? Or is KS looking for a rational approach? I personally was flummoxed by this question - hardly unusual, especially on this forum.

Not quite hunt and peck but not very sophisticated either!

First, I thought that the number must be round. By round

I mean that it has a large number of factors given its

size. 30 is a round number by this definition. So, I

wrote a program to go up from some small number like 10

and print out numbers which had *lots* of factors. Well,

there were so many that I decided to modify the program

to take all subsets of 8 factors and add them up to see if

any added up to 3240. The first number that did this was

1260. In fact, I found that 1260 has so many factors that

you can find subsets of 8 which add to any number between

3239 and 3243, and is also the smallest number having these

sums for subsets of 8. I'm ashamed to say that it was

brute force. I hope KS clues us in if there is a way to

do it which is less shameful!

1260. Since 3240 = 1260 + 63 + 90 + 210 + 252 + 315 + 420 + 630

Given that x <= y <= z

xy + xz + yz <= 3zy

rearranging: xy + yz + zx - xyz = 2

to get: xy + yz + zx = 2 - xyz

and the substitute: 2 - xyz <= 3zy

(2/zy) + x <= 3

From this we can see that x must be less than 3. I can't put an upper limit on y or z yet.

Thanks for the start, psychic-mind. As you

have showed, x=1 or 2. If x=1, we plop it

into the original formula and get,

y+z+yz-yz=2. So, we have to solve

y+z=2 in the positive integers. Hence, y=z=1.

Hence, the solution (1,1,1).

Now, when x=2, we get 2y+2z+yz-2yz=2 which

is 2y+2z-yz=2. Solving for z, we get

z=(2y-2)/(y-2). Now, y<=z so

y<=(2y-2)/(y-2) from which we get

y²-4y+2<=0. This is true when

2-sqrt(2)<=y<=2+sqrt(2). The only positive

integer greater than x in this range is 3. Plugging x=2

and y=3 into the original, we get z=4.

Hence, the (only) other solution (2,3,4).

"Nostalgia is a longing for the place you would never dream of moving back to"

I knew someone would solve it! Nice (but difficult) going! That wraps up the Walter Penney puzzles. You guys solved all of them.

I believe that these two answers comprise the entire solution.

So do I, but I'm having difficulty proving it.

(1,1,1). Beyond that, "xy + yz + xz" grows faster than "xyz" so the difference will be more than 2.

(2,3,4) ?

Certainly understandable if you do not care to answer... Are the two halves of the message split between words or within one word?

They are split within a word.