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Everything posted by jim

  1. 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.
  2. Saxguy....Realistically you are correct--but in the original version of the problem it was specifically stated that both killers die when they meet each other.
  3. jim


    Could the answer be 0 bats--since nothing was said in the subsequent lines about the bats mentioned being in the cave?
  4. 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.
  5. B for men, B for women, so obviously B is the better treatment regardless of gender--although this might not be statistically significant. But if you had only told me the overall numbers without a gender breakdown I would have picked A--but given the gender breakdown info I pick B.
  6. Bobby Go . Nice answer. I missed 270. There is a general method to find the largest such number with 3n-1, 3n, or 3n+1 as the number of ones--where n= m+1. It is easiest for 3n.
  7. 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?
  8. EXTENSION PROBLEM. What is the smallest number that requires at least 16 ones--and what is the largest number that can be expressed with no more than 16 ones?
  9. 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.
  10. 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.
  11. 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?
  12. 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?
  13. 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.
  14. EXTENSION 2--Same 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 balls--but it is easy to generalize to k blue balls, just a little messy to write up the answer.
  15. jim

    Plane crash

    I'm sure the people in Canada would say the US because Canadians are too civilized to bury people alive,
  16. EXTENSION PROBLEM--only 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.
  17. 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 die--otherwise it counts as survival. Now do as before. It may be possible to do with less than 18, but i would not no how.
  18. 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.
  19. bonanova would be correct if there were 1000 bottles.
  20. 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.
  21. It seems very unlikely that dart players in a pub would play a game where they simply count the numner of bullseyes.
  22. 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 m-2 is at least 3 and as a smaller number greater than 2 we can do it. We adjust the solution for m-2 by increasing one of the odd numbers by two, replacing it with an odd nuimber not already included in the set of m-2. Since the solution for m-2 is not the null set, we can always do this unless the solution for m-2 is of the form U(j) which is all the allowable odd numbers from j upwards for some j less than or equal to 2n-1. But in this case we can replace j with 1 and 3 and j-2 which we can always do as long as j>5. If j=5 we get our sum m-2 = n^2 - 4 so m = n^2 -2 which is in fact the smallest non-solution except for 2. j=3 and j=1 give us m-2 = 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.
  23. 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 2n-1 except for the numbers 2 and n squared - 2
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