superprismatic

I'm a bit confused because I don't know what you mean by symmetries. Would you mind defining symmetries?

Sure

I ignored symmetries as you had not stated that they are not to be ignored. Ambiguous phrasing is not the rule in mathematics and logic, which is what this forum is about. Clarity is important unless loosened parameters are stated up front. By the way, why did you ever thing that I thought "Omni-" meant "law"?

Let's take the cube calculation. There are two possible diagonals for each face. There are 6 faces. So, ignoring symmetries, there are 26 possibilities for drawing diagonals. If you are asking more than this you should state it unambiguously.

Why?

1. 64 not counting symmetries 2. Two, because it's the only way "not further subdivided" 3. Empty 4. Omni-

Join the club. I haven't been able to prove it either. I'm glad you enjoy the puzzle. Thanks for the feedback!

But, there are 9s in the grid. So, the numbers can't be all mod 9. Confusing!

I just wanted to point out the fairly remarkable coincidence that the first row of Pickett's updated chart (8,7,?,4,1,?,?,2,?) agrees with the fibonacci sequence taken mod 9 (the first row values are in bold): 0,1,1,2,3,5,8,4,3,7,1,8,0,8,8,7,6,4,1,5,6,2,8,...

I expect that Pickett looked at the #7 post and googled Hulahup cache. When you do this you get the original cache in French as the third choice.

Nice problem, nithinroks! Thanks.

Nice! I wasn't sure how many solutions there were. I was hoping that it was unique because I thought there were enough constraints to do that. After I made up the problem, I solved it by hand. I never wrote a program to look for answers.

Like witzar, I had thought that the Kelly Criterion gives the largest expected value of the final bankroll. It actually maximizes the expected growth of the bankroll. So, in the case of this topic, if p is the proportion of the bankroll to be bet each time, the expected bankroll increase for each play is: (1+.9×p).55×(1-p).45 which attains its maximum when p=.05. So, I was wrong in my answer (#6) to maurice. In short I was looking for the answer to a different problem than the one I posed. Sorry about that, especially to maurice! Maurice has the correct answer to the problem I posed -- the let-it-ride strategy does indeed maximize the expected value of the final bankroll.

Here's something I came across in my mathematical perusals: 0 1 1 2 3 5 8 13 ... 1 3 4 7 11 18 29 47 ... 2 4 6 10 16 26 42 68 ... ...etc... [/code] Notice that each row above is fibonacci in that, except for the first two columns, a number in the i[sup]th[/sup] column is the sum of the numbers in the (i-1)[sup]st[/sup] and (i-2)[sup]nd[/sup] columns. Also notice that the first column is just the one-up integers starting at 0. Here's how to generate the rest of the array: Label the columns -1,0,1,2,3,... The first row is the usual fibonacci sequence. We will genetate the rows consecutively beginning 1, 2, 3, etc. For the row beginning with N, we look for N in columns labeled with [i]positive[/i] integers and in the rows above the one we are generating. When we find N in this manner, we take the number immediately to its right, add 1 to it, and put this next to N in the line we are generating. Then we generate the rest of the row using the fibonacci rule For example, to genetate the row beginning with 1, we find the 1 in the column labeled 1 and take the 2 immediately to the right of it, add 1 to it to get 3. That 3 becomes the number next to the 1 in the line we are generating. Another example: The row which begins 3. We find a 3 in the top row with a 5 next to it. We add 1 to the 5 to get the 6 as the element in the 0[sup]th[/sup] column of the row beginning with 3. Using the fibonacci rule, we find that row to be: [code] 3 6 9 15 24 39 63 102 ... The claim was made that every positive integer can be found in one, and only one, positive integer labeled column and one, and only one, row of this array. Can you prove this?

Right? : I didn't do any calculations for only a small number of days. So I don't know the N for which, after N or more days, the best strategy is unique. The let-it-ride strategy will make you broke in short order, so the best strategy would need to bet some fraction (which may or may not change) of your money each time. Anyway, feel free to use 365 days as "the long run". I'm sure that the best strategy doesn't change from that point on.

You have a biased coin which falls heads 55% of the time and tails 45% of the time. You start with $1 and, once a day, you may bet any fraction of your current amount that the coin will land heads. If it lands tails, you lose the amount that you bet; if it lands heads you get 190% of your bet back. So, for example, say on some day you have $150 and decide to bet $10 of it that day. If the coin lands tails, you lose your bet and then have $140 available for the next day. If it lands heads, you win $9 plus your $10 bet back (190% of $10) and then have $159 available for the next day. What is the optimal betting strategy? That is, what strategy will give you the largest expected value in the long run? Please Note: For simplicity, assume that money values are real numbers. So, for example, one may have $124.60012898735.

Indeed, you are correct. I had erred in my problem statement. I meant to have the puzzle have the feel of CrayolaSunset's problems in that different letters are assigned different number values. This is what makes the problem difficult. Sorry. I will amend the problem statement.

*** Inspired by CrayolaSunset's nice puzzles *** For my 1,000th post, here's one that I think is really hard because I think it requires several insights which don't seem to pop to mind (at least not mine): The letters of the alphabet used in this puzzle were each assigned a unique integer in the range 2 to 26 such that no letter is assigned a value of a prime larger than 13. Words are given values that are the product of the values assigned to the letters. For example, if P=2, A=3, and T=5, then PAT = 30. If FINK - (AQUEOUS)(1/4) - (PILLS)(1/3) - (BUTTER)(1/2) = 154 (roots are all positive integers), then what are the values of FINK, AQUEOUS, PILLS, and BUTTER?

Yep, that is just silly. But that's part of the fun here at the Den. We're serious about our silliness here!

Well, the value of the letters f,h,q, and v can't be determined. But there are 4 sets of values for these which are possible (see my spoiler in post #2). None of these letters are in “using your mind”, so they don't matter for the solution to the problem.

I would call this puzzle an Alphamultiplic Puzzle. I made that up. I'd like to suggest showing the prime factorization of the values in order to make it easier on solvers. In this case the clues would look like: Gamble = 1,297,296 = 24·34·7·11·13 Roulette = 917,280 = 25·32·5·72·13 ...etc... But this is your baby, so it's entirely up to you.

I usually look at the latest publications from the Mathematical Association of America (MAA) which can be found here and order interesting-looking books designated Dolciani or Problem Books. I order them through my local county library. By the time I am just about done leafing through those, a new MAA publications catalog comes out with more good books. I don't usually record in my notes what book the ideas came from, so I don't have a list of books immediately at hand.

How do I come up with my problems? Well, I like to keep a few relatively new expository mathematics texts nearby so that I can fill some of the time slots in my life. I routinely keep notes as I flip the pages. These notes are usually the basis of my posted topics. Something that surprises me, like the odd binomial coefficient thing, usually makes it in one of my topics. So, most of my topics are not original, but some are. I find my perusal of expository texts gives me lots of insights. I usually get these books using inter-library loan from university libraries. I'm glad you like some of this stuff, Captain! Thanks!

I mean that nCk is odd if and only if the binary representation of n contains a 1 in every position where the binary representation of k contains a 1. So, for example, if n were 84, 1010100 in binary, then the only values of k which would make 84Ck odd is 0,4,16,20,64,68,80 and 84 (binary 0000000, 0000100, 0010000, 0010100, 1000000, 1000100, 1010000, and 1010100). In terms of the way I said it in the OP, 84C20 is odd because 20 = 84 AND 20 (0010100 = 1010100 AND 0010100). I should have used parentheses in the OP to make it k = (n AND k). I hope this makes it clear.