jim

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Saxguy....Realistically you are correctbut in the original version of the problem it was specifically stated that both killers die when they meet each other.

As long as the process reaches completition within the yearor as long as the probability of A meeting B in a short interval of time does not depend on how many other people there are (a kind of Ideal Gas Law) your chances of survival are the same whether you chose to be a killer or a pacifist. That is because the only encounters that matter for you are when two killers meet, or when a killer meets you. Whether a pacifist dies when meeting you are not is irrelevant to your survival.

Could the answer be 0 batssince nothing was said in the subsequent lines about the bats mentioned being in the cave?

Perhaps this will help clarify things. Suppose we find the average height of college men is greater than it was 60 years ago, the average height of college women is also greater, but the average height of college students is less than it was 60 years ago. Would this suggest that people are taller or shorter than they were 60 years ago.

B for men, B for women, so obviously B is the better treatment regardless of genderalthough this might not be statistically significant.
But if you had only told me the overall numbers without a gender breakdown I would have picked Abut given the gender breakdown info I pick B.

Bobby Go . Nice answer. I missed 270. There is a general method to find the largest such number with 3n1, 3n, or 3n+1 as the number of oneswhere n= m+1. It is easiest for 3n.

Comments. The smallest number does require a lot of messy testing. But the largest number is another matter. As noted above, we can do 256. There are also two larger numbers we can also do. What are they?

EXTENSION PROBLEM. What is the smallest number that requires at least 16 onesand what is the largest number that can be expressed with no more than 16 ones?

You don't need calculus. We just need to know the area of the part of the solid at least t / k cm away from an exposed surface. The radius decreases at the rate of t / k cm per second. The upper surface drops at the same rate, and so does the bottom surface except in the case where the bottom is not an exposed surface. As long as the height is at least 2R(or R if the bottom is not exposed) then the radius decreases linearly down to zero. Otherwise the height goes to zero while the radius approaches a positive limit.
In your first point the limit is a point. With height 3R it approaches a limit that is a line of height R. If the height is R it takes half the time to disappear, and the limit is a 2D disc of radius R, etc.

You would need to know the initial radius R. Simplifying your rate you get k cm/s as the rate at which the radius decreases so it would take 0.5R/k to get to half the radius and R/k to sublime completely.

You can answer this question if you solve a single problem: What is the probability of meeting if one friend waits 15 + x minutes and the other waits 15  x minutes?

Question. If we start with 45 piles of one card each, do we immediately destroy all piles and finish with a configuation of no cards and no piles in no moves?

Any real number can be expressed by a sign, a finite number of binary digits before a binary point, and then a possibly iinfinite and definitely countable number of binary digits after the binary point. 0.1010010001... would be an example of an irrational number in binary where every run of 0"s countains one more 0 than the previous run and is followed by a single 1.

EXTENSION 2Same problem but now assume we add two balls red with probability r and blue with probability b. What about when we add three balls? This time it will we convienent to assume we start with 10 blue ballsbut it is easy to generalize to k blue balls, just a little messy to write up the answer.

I'm sure the people in Canada would say the US because Canadians are too civilized to bury people alive,

EXTENSION PROBLEMonly slightly more difficult. Instead of assuming the added ball had an original probabilty of 1/2 red and 1/2 blue assume a probability of r red and b blue, with r+b = 1 of course.

I flip the coin twice.
HT is one outcome and TH is the other. If I get either HH or TT I repeat the procdure.
 I look for the first number in Pascal's triangle that is at least N. I select N possible outcomes with the same number of heads and tails specifying which correspond to j/N and which to (Nj)/N. In your puzzle j=1 but it could be any number less than N and relatively prime to N. A success is getting one of these outcomes.
Following a failure we repeat the process with a difference. For our first try our best guess for the ratio of the likelyhood of getting head and tails was 1:1. Now it would be the maximum likelyhood estimate of (h+1):(t+1) where h is the total number of heads we have gotten up to this time and t is the total number of tails. This time we select a number from Pascal's triangle where the ratio of the likelyhood af a successful trial to the number of flips involved is as large as possible

You can do it with 18. In place of each prisoner in the solution to the previous problem, use a pair. It counts as dying only if both prisoners dieotherwise it counts as survival. Now do as before. It may be possible to do with less than 18, but i would not no how.

bonanova As a math teacher I considered your solution to be correct except for one minor detail. As far as I am concerned you are the one who solved the problem.

bonanova would be correct if there were 1000 bottles.

Suppose they each throw 100 times with 97 bullseyes for Alex while Davey gets 90 and Ian gets 80. What would the team of Davey and Ian score? Suppose Davey throws first and misses 10 times. Ian lets the pressure get to him and misses every time Davye does.lus 10 additional times. In this case the team scores 90. On the other hand, if Davey is good under pressure and makes all 10 of those throws while missing on 20 other occassions, then the team scores 100. If the correlation coefficent for Davey and Ian is zero, then the team scores 98 and they win. With a negative correlation they do even better, but with a large enoiugh positive correlation the bet will be a good onwe for Alex.

It seems very unlikely that dart players in a pub would play a game where they simply count the numner of bullseyes.

Event Horizon
I think your proof is essentially the same as mine, but formulated in a different fashion.
I basically used induction. We can do 3=3 and 4=3+1. Let m be the smallest number greater than 2 which we cannot do. It must be at least 5 so m2 is at least 3 and as a smaller number greater than 2 we can do it. We adjust the solution for m2 by increasing one of the odd numbers by two, replacing it with an odd nuimber not already included in the set of m2. Since the solution for m2 is not the null set, we can always do this unless the solution for m2 is of the form U(j) which is all the allowable odd numbers from j upwards for some j less than or equal to 2n1. But in this case we can replace j with 1 and 3 and j2 which we can always do as long as j>5. If j=5 we get our sum m2 = n^2  4 so m = n^2 2 which is in fact the smallest nonsolution except for 2. j=3 and j=1 give us m2 = n^2 1 and n^2 which in either case give us m> n^2 which we can also not do because they are too large.

I would like to propose an extension of this problem. Same conditions. Show every number from one to n squared can be expressed as a sum of distinct odd integers from 1 to 2n1 except for the numbers 2 and n squared  2
Infinite monotone subsequence
in New Logic/Math Puzzles
Posted
You need to change to the standard definition of montone. Monotone increasing means each number is at least as large as the previous number. If each successive number is actually larger that is called strictly monotone. Decreasing monotone is defined in a similar way. Note 1,1,1,1,1,1,1,... would need my definition to have any montone sequence of length greater than one.