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  1. Yesterday
  2. For the second version, with the 1g precision scale, it is possible to go up to 41610 cases and not only 41010. I forgot one trick. sorry about that.
  3. Last week
  4. @CaptainEd : there is some idea... I just remind you that you can use the scales only two times each!
  5. Hi, this is a riddle I just created. Sorry for my broken English, it's not my native language. There are two version of almost the same idea. The 1st one is purest, but the second one is more tricky. I think it's quite difficult to solve, but there is no need for any mathemical knowledge. First version of the riddle A poker manufacturer has created 10201 cases each containing 100 chips. (Note: 10201=101*101) 10200 of these cases are perfect and each of their chips weighs 10g. One of these cases is defective and its 100 chips weigh 11g each. In order to determine which of the 10201 cases is defective, the manufacturer has two scales at his disposal: A traditional, high-precision scale that can hold tons. An electronic kitchen scale that displays the weight with a precision of 2g and can hold no more than 100g. How can the manufacturer determine the defective case using only 2 times each of the 2 scales? Second version of the riddle A poker manufacturer has created 41010 cases each containing 100 chips. 41009 of these cases are perfect and each of their chips weighs 10g. One of these cases is defective and its 100 chips weigh 11g each. In order to determine which of the 41010 cases is defective, the manufacturer has two scales at his disposal: A traditional, high-precision scale that can hold tons. An electronic kitchen scale that displays the weight with a precision of 1g and can hold no more than 100g. How can the manufacturer determine the defective case using only 2 times each of the 2 scales?
  6. Earlier
  7. Thanks ! We can see the days we need is sumi<=P+1ap<3*ap+2=3*43*ap-1<3*43*6P, which is O(P).
  8. You got it. Well done. I'll post my solutions later.
  9. To solve the question without knowing parity, we follow almost the same steps, except we change hypothesis of parity alternatively each time we go to next round. And it may take us one more round to find it, because we find it only when in the same parity.
  10. I tried again like this, supposing we know the parity case, for example odd on first day By checking (1,3),(4,6),(7,9),..., we can check a range of (3k-1) in k days. Now I construct a sequence of numbers ai, such that an >= 3*sumi<nai we can easily verify that ai = 4i qualifies. The strategy now becomes, in the first a1=4 days, we check (-5,-3),(-2,0),(1,3),(4,6), which is a range of -5 -> 6 in the next a2 = 16 days, we check (-23, -21),(-20,-18),...(-2,0),(1,3),..,(22,24), which is a range of -23 -> 24 etc. Supposing groundhog is at P, while |6P|<ap = 4p Then after ( a1 + a2 + .. +ap ) days, the range we just checked is the range of last ap days, which is about -1.5*ap -> 1.5*ap, while the range of groundhog could be is -|P| - ( a1 + a2 + .. +ap ) -> |P|+( a1 + a2 + .. +ap ), on the other hand, we have |P|+( a1 + a2 + .. +ap ) - 1.5*ap = |P|+( a1 + a2 + .. +ap-1) - 0.5*ap < ap/6 +ap/3-0.5*ap = 0, which is to say, groundhog's range is included by our last searching range.
  11. Apologise I considered it to be 2 days after when they come back to positive half axis, yeah, after fixing this 4K hole doesn't work for 4k day. Is this an open question or you know there's an answer ?
  12. Unfortunately, that does not work for the known parity case. Notice day 4 checks even holes, and day 5 also checks even holes. So days 5-8 are checking the wrong parity. Fixing this, the numbers change such that day 4k+x checks hole 4k+y, so the groundhog can escape by fleeing away from hole 0.
  13. She would say “You are going to eat my child”. In order to make the sentence true the crocodile would have to eat the child. But he said if she guessed right then she could have the child back. But if he gave the child back, she would be wrong because he didn’t eat the baby, so he would have to eat the baby. But since he ate the baby, she would be right so he would have to give the baby back. If she used the sentence it is a paradox because no matter what happens it will always contradict the outcome every single time
  14. xp2008

    For 1st question, we can't guarantee to find it. Supposing groundhog knows your strategy, thus knows which hole you check next day, it can always avoid it because it has always 2 choices for next's hole.
  15. If we know first day's parity then we can. Supposing it was odd, then I will check 1 3 4 6 1 -1 -2 -4 6 8 9 11 ... Which is to say, day 4k+1, I check 5K+1, 5K+3 day 4k+2, I check 5K+4, 5K+6 day 4K+3, I check -5K+1, -5K-1 day 4K+4, I check -5K-2, -5K-4 We can see my range is ( -5K-4, 5K+6 ) , while lim ( 5K - 4K) = lim (K) -> infinity, thus we could always find the groundhog in this case.
  16. The sequences are different in each grid, can you solve all 4 question marks? Grid 1: Grid 2: Note: Since it are 2 seperate grids, the sequence logic is different.
  17. There is no end to start from. Every hole has an infinite number of holes on both sides of it. And, yes, both you and the groundhog are immortal. Your friend, not so much. Luckily, you'll always be able to enlist a member of his posterity to help check one hole each day, so close enough.
  18. I think I can do better: But not sure it is the max.
  19. If they started from either end worked to the middle they would eventually find it... provided they lived for eternity
  20. You get a pain in the arse
  21. If you sit on a tack, you don’t take a aspirin -what does this mean?
  22. Hello, I have one Math puzzle to which I think I found the solution but wanted to check your opinion on it. The puzzle is the following: George got a birthday present which was 1 banknote of 100 (the currency does not matter. Lets name it "currency" just for convinience). Valid banknotes in the currency are: 100, 50, 20, 10, 5, 2 and 1. So he decided to spent some of the money in the shopping mall (he did not take any other banknotes with him. Just the one banknote of 100). At the end of the shopping it turns out that: 1. In each and every shop he bought just one item and therefore just one payment was made. 2. The price of the item was a whole integer (no decimal points are allowed) 3. For each item he never had the exact sum that's why he always gives the nearest banknote (of which he had at the moment). 4. The sellers on the other hand always have enough money in different banknotes that's why they return the change with as less banknotes as possible. However, at the end it turns out that each seller aways return at least two banknotes to George. The question is: What is the maximum number of items that George can buy with these restrictions? I have bolded the important parts in my opinion, so they pop out. The approach that I took was the following: 1. In each shop the way I tried to figure out the price of the item is to have a change with as less as possible banknotes as change and with as much as higher value of the banknotes in the change 2. And as less as the price can be. So the solution: 1. First Shop one buying of item that costs - 10 - The payment was made with the available banknote - 100 - The change was - 50 20 20 - Available banknotes for George - 50 20 20 = 90 2. Second Shop one buying of item that costs - 9 - The payment was made with the available banknote - 20 - The change was - 10 1 - Available banknotes for George - 50 20 10 1 = 81 3. Third Shop one buying of item that costs - 7 - The payment was made with the available banknote - 10 - The change was - 2 1 - Available banknotes for George - 50 20 2 1 1 = 74 4. Fourth Shop one buying of item that costs - 9 - The payment was made with the available banknote - 20 - The change was - 10 1 - Available banknotes for George - 50 10 2 1 1 1 = 65 5. Fifth Shop one buying of item that costs - 7 - The payment was made with the available banknote - 10 - The change was - 2 1 - Available banknotes for George - 50 2 2 1 1 1 1 = 58 6. Sixth Shop one buying of item that costs - 9 - The payment was made with the available banknote - 50 - The change was - 20 20 1 - Available banknotes for George - 20 20 2 2 1 1 1 1 1 = 49 7. Seventh Shop one buying of item that costs - 13 - The payment was made with the available banknote - 20 - The change was - 5 2 - Available banknotes for George - 20 5 2 2 2 1 1 1 1 1 = 36 8. Eight Shop one buying of item that costs - 17 - The payment was made with the available banknote - 20 - The change was - 2 1 - Available banknotes for George - 5 2 2 2 2 1 1 1 1 1 1 = 19 I tried different approaches and this gives me the highest number of items - 8. For the last three we have several options but they cannot give more than 8. Is this the correct answer and if no how can you approach the problem to have more buyings. I found the rule that sellers should return at least two banknotes the most restrictive.
  23. Perhaps this question will help move things along...
  24. Can anyone find the groundhog using less than 5 hunters?
  25. Barring blind luck, I don't think it can be caught with two hunters. Thoughts....
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