Prime

What is the base for the logarithm? It makes sense to consider base x. Let us also designate the infinite string of x exponents as p(x). The condition is that p(x) =2. Then logxp(x) = logx2 p(x)logxx = logx2 p(x) = logx2 2 = logx2 And thus x = 21/2 Having that, I am not sure my answer for the p(x) = 4 in my previous post is correct.

I don't see 21/2 as an answer here. To me it seems more like 2^(21/2). Inductive reasoning could be the same as in my previous post.

That depends...

Here is my inductive reasoning. I don't know if it could be considered as a proof, though.

I already tried to prove the same point to Bonanova. But somehow, he went ahead and won a gambling match against me.

I understand. However, formal proof is not necessarily the same as common sense. You say that we roll six times and each number appears once (on average) because all numbers equal and we should not give a preference to any one of them over the other. That is an obvious sensible view, but I don't see it as a proof. I would be more inclined to accept that line of reasoning as a proof if you showed that all other possible views lead to a contradiction. E.g., Suppose some numbers are more equal than others. Then ...

I type too slow and people manage to make several posts while I work on one. So I'll repeat what I said in my previous post. I think you use a recursion here. "Six trials, six equally likely results" is a presumption of what the average is. The consecutive reasoning is based entirely on that presumption. It is not a proof. Such reasoning is simply a demonstration that if the average was what we presumed it was (1/p) that would lead to no contradiction.

That is an intuitive approach that looks convincing enough. But as a proof it seems recursive. Basically, what it says is: Because we know that the average is this, we can show that it is indeed this. As for the corollary that was the very thing that we were proving here. Y-s proof is the one we can't argue with since it makes no presumption of what average is, but establishes it from the initial conditions (probabilities). However, there are simpler "cuter" proofs, although may be also questionable. For example: The average of an event is its probability times its weight (number of occurences). Say probabiblity of an event to occur on a single try is "p" (known) and we designate the average as "a" (an unknown to be derived). We can view the average as a sum (probability times number of occurences) of the event occuring on the first try, and not occurring on the first try. Then the average may be expressed as following: a = p*1 + (1-p)*(1+a) (If it did not occur on the first try, now we expect it should take a+1 tries.) --> a = p + (1-p) + (1-p)a --> a = 1/p.

Yes, if...

This has been done recently in here.

Vimil has the exact number! Imran came close, but missed the powers of 5.

Sorry, that's incorrect. I did not mean for anyone to use calculator or computer program to count the actual number of zeroes at the end. I meant to find the answer analytically. Use calculator only for what fits entirely on the display without "E" notation. As you have noticed:

Even if your calculator could come up with exact decimal representation of 2008!, it would take you several minutes to count the zeroes at the end.

Just a bit more challenging factorial problem. How many consecutive zeroes does the number 2008! end with?

Perhaps...

This problem is a much smaller cousin of the unsolved square division on this forum. Interesting how minimization requirement is so much easier than total count problem. Here are some essential elements for proof of the 280 5/13 as an absolute minimum.

With respect to the longest cuts, I suggest we should assume each cut to be one straight line cutting all the way across. (Everything into two pieces.) But let's return to the lower limit. I think the theoretical 7 cuts 275cm proven by Bonanova may not be the highest lower limit for dividing 1m square into 100 equal parts. Here is a set of 7 cuts adding up to 280 5/13 cm, which I think is the highest lower limit. (It is not possible to make total cuts shorter.)

I don't see why you are backing off from 299.5 cm. Here is a set of simple straightforward cuts adding up to it. The interesting problem is to prove what's the absolute minimum...

How did you know when to stop cutting and burning? Although, I am beginning to see your point. What you get is as close approximation to 1/n of the total time as you can manage by moving fast. Unless, you got lucky at some point and divided remainders into n of equal time length. Here is how I would express it in general terms. Lets take an example of n=3. That is we want to measure T/3 time interval using uneven cord of total burning time T. Cut the cord into 3 arbitrary pieces and ignite them at the same instance. Suppose the fastest took time x1 to burn. Then the total worth of burned cord is 3x1 at the instance it finished. Cut quickly an end from the longer remaining piece and ignite it. Again you have 3 pieces burning simultaneously. Suppose the fastest took x2 time to burn. Thus you have burned another 3x2 worth of cord and thus far x1 + x2 total time elapsed. Repeat the procedurre ad infinitum. In the end you have burned the entire T = 3x1 + 3x2 + ... worth of the cord. While the total time elapsed is x1 + x2 + ... Which happens to be exactly 1/3 of the total time. Interesting! Here is the procedure for simultaneous ignition and division: When you see one of the cords is about to expire, cut off the end of another cord and join it to the non-burning end of the expiring cord. This way you don't need to be all that fast.

I can't resist: "Your view is blocked by Bonanova's head." Or did you mean CR approaches you from behind?

Work out an example. Say you want to measure 10 min. Take one of the cut pieces burns in 2 min, second in 4, and the third in 24 (or one half of that time each when lighted from both ends.) Or use your own numbers. Remember, when you cut any piece in half lengthwise, it does not tell you how fast each half would burn.

I don't seek recognition. Especially, for dazzling solutions, which are not my own. I just enjoy solving problems. And for that matter, I disagree with posting answers from books and other websites. At any rate, it should have been up to Y-s to post a book answer on this topic. Which, as I looked it up, is dn = 1/(1 - cos(2pi/n)), where n is the number of sides of a regular polygon. There are different schools of thought on how to ejnoy riddles. Some people like to look at the problem for a few minutes and then head for the answers section. Others only want to find solution on their own. Both ways are perfectly legitimate. Unfortunately, there is a conflict of interest there at times. Logically, I don't see any reason for posting solutions which you have found in literature. Those who want to find the answer can do the same. Also, note that people come upon posts at different times. And it would be interesting to someone who stumbled upon this problem few months from now and found it was still unsolved by the forum.

I second that. I interpreted "... as I had" in the "I would have ... as I had" clause as the actual distance traveled -- not as the distance remaining. In other words: "If I had gone 3 times as far as I had, I would have half as far to go as I had gone"

For the adjacent corners it seems to be the case. There are four identical non-intersecting areas covering the entire square. My initial instinct tends towards more complex. But since Y-s put Prime and Bonanova into adjacent corners, your interpretation must be the true one.

That's an awesome solution by Bonanova. However, I'll hold back on praising for now, on the account of the "related" problem. To me that's an indication that Bonanova already knew the problem and/or read about it. If that's the case, he simply denied our chance of solving it. If it is not the case, I appologize. The solution is really ingeneous. With respect to the area sweeped, I don't see how it is 1/4. The way I see it...