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CaptainEd

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Posts posted by CaptainEd

  1. This is similar to the proofreaders problem.


    Yes, here’s why.

    Assume the stonenibblers’ probabilities in a time interval are p and q, and they are independent events.
    Then the probability of both nibbling in the same interval is pq.

    The probability of only one in a time interval is p(1-q) + q(1-p).

    The probability of no sound in an interval is (1-p)(1-q).

    So we have three equations

    (A) pq = .17

    (B) p-pq + q-qp = p+q-2pq = .38

    (C) 1-q-p+pq = .45

    Substitute pq in (b) and (c)

    (B) p+q -2*.17 = .38 p+q = .72

    (C) 1-(p+q) + .17 = .45

    1-.72+.17 =.45
    .45=.45

    So it does look like their noises occur independently.

    [\spoiler]
     

    [\spoiler]

  2. Here’s one way, there may be a quicker one.

     


    start at 1, open them one at a time up through 10. Then hit 10 again, and go back one by one to 1.

    Heres my argument:

     

    The bug is in one of two states: 
    (a) box number same parity as time number, such as box 3 at time 1, or box 2 at time 4, OR

    (b) box number opposite parity as time number, such as box 4 at time 1, or box 2 at time 3

    The proposed approach is to spend the first 10 times in state (a), then, by taking two time units in box 10, we switch to state (b) and come back to 1.

    If he was in state (a), then we would have found him on the forward pass, but if he was in state (b) we will find him in the backward pass.

    For example, if he is in box 2 at time 1 (state b) when we open box 1, we miss him. At time 2 he could move to box 1 or 3, and we miss him again. By time 10, when we are opening box 10, he can be in any odd box. For example, he might be in 9. At time 11, we open 10 again. The bug either moves to 10 (and we catch him) or he moves to 8. So we move to 9 and he moves either to 9 ( and we catch him), or to 7. As we continue back, we are always an even distance from the bug, including 0. This can continue until we open 4 while he was in 2. When we move to 3 he either moves to 3 (and we catch him), or he moves to 1. In that case, we move to 2 and he has to move to 2 (and we catch him).

    On the other hand, if he is in State (a) when we start (also in state a), then we are an even distance from him, and we will catch him by time 10.


    There was a brainden puzzle with a video that reminds me of this puzzle. Look for “Fox in a Hole”)

     

     

  3. I assume that they also all get the clue that their numbers are unique.


    the other numbers are 37 and 3*37= 111

    Here’s the argument:


    If the numbers are n, 2n, and 3n, then the person with 3n would recognize that fact because seeing n and 2n, the only choices would be n and 3n, but since n would not be unique, it would have to be 3n.

    So the numbers are not n, 2n, and 3n.

    Therefore, if someone sees n and 3n, they know they don’t have 2n, or the 3n person would have recognized it. So they must have 4n.

    So the person with 148 could be 4n, seeing 37 and 111.

    Neither of the other two could rule out their two possibilities.


     

    • Like 2
  4. Thank you Rocdocmac. Once I’m done with this, I’m going to take a high level math class on how to divide by 4.

     

     


    Birthday 3/12/2000

     

    Analysis follows:


     

     


    first number: the last digit of powers of 2 goes through the cycle (2486), with an exponent  divisible by 4 ending in 6. The given exponent 667788 is divisible by 4, so this number is 6. 

    second number: the last digit of powers of 3 runs through the cycle (3971), with exponent divisible by 4 ending in 1. 556677 is 1mod4 so it ends in 3.

    third number: the last digit of powers of 7 runs through cycle (7931). The exponent 445566 is 2mod4 so this number is 9.

    fourth number: last digit of powers of 8 has cycle (8426), and the exponent of 334455 is 3mod4, so this number is 2


    So age = 6 + 3 + 9 + 2 = 20

     

    I hope that’s that!

  5. How embarrassing! I see your point, Rocdocmac. Here’s the new improved answer:


    first number: The given exponent 667788 is divisible by 4, so this number is 6.
    second number: 556677 is 3mod4, so it ends in 7.

    (Improved) Third number: The exponent 445566 is 2mod4, so this number is 9

    (improved) fourth number: the exponent of 334455 is 1mod4, so this number is 8.

    Total = 30

    birthday mar 12, 1990

  6. I’m pretty sure I know; the answer follows and then hidden analysis:

    Andy’s birthday was 3/12/2004

    analysis follows:


    first number: the last digit of powers of 2 goes through the cycle (2486), with an exponent  divisible by 4 endinG in 6. The given exponent 667788 is divisible by 4, so this number is 6.

    second number: the last digit of powers of 3 runs through the cycle (3971), with exponent divisible by 4 ending in 1. 556677 is 3mod4, so it ends in 7.

    third number: the last digit of powers of 7 runs through cycle (7931). The exponent 445566 is divisible by 4, so this number is 1.

    fourth number: last digit of powers of 8 has cycle (8426), and the exponent of 334455 is 3 mod4, so this number is 2.

    His age, on his birthday 3/12/2020, was sum(6,7,1,2) = 16
    so if we just count years and not distinguish leap years, he was born 3/12/2004.


  7. Event Horizon, you have shed a lot of light on this puzzle! Your observations about the 4-triangle and  1+ 3^n are quite surprising and pleasing. It is I who brute forced it; you actually came to a deeper understanding. I’ll send you kudos as well.

  8. Answer and proof:


    First, think of the colors as ternary numbers, 0, 1 and 2

    The rules are

    00->0

    01->2

    02->1, 

    11->1, 

    12->0, 

    22->2

    Notice that this function simplifies to
    XY->(3-X-Y)mod3 = ( -(X+Y)) mod3

    Specific case: XX-> -Xmod3

    Let’s name the blocks A through J, and use a non-standard notation:

    AB-CD means (A+B-(C+D))mod3

    The blocks are separated by semicolons.

    (1) A;B;C;D;E;F;G;H;I;J

    (2) -AB;-BC;-CD;-DE;-EF;-FG;-GH;-HI;-IJ

    (3) AC-B;BD-C;CE-D;DF-E;EG-F;FH-I;GI-J;HJ-I

    (4) -AD;-BE;-CF;-DG;-EH;-FI;-GJ

    (5) ABDE;BCEF;CDFG;DEGH;EFHI;FGIJ

    (6) BE-ACDF;CF-BDEG;DG-CEFH;EH-DFGI;FI-EGHJ

    (7) AG-D;BH-E;CI-F;DJ-G

    (8) DE-ABGH;EF-BCHI;FG-CDIJ

    (9) ACEGI-BDFH;BDFHJ-CEGI

    (10) -AJ

    So the method is: discard the 8 middle blocks, and calculate using the original rule on the first and last blocks.

    • Like 1
  9. How did all those 2/11 become Thalia’s answer?

    draw a 6x6 table, and tally one in each cell each time Plato asks about another number (11 cells per number). When we are done, all 6 of the diagonal elements (11,22,33,44,55,66,) have one tally each, while all of the off diagonal cells have two tallies. Therefore, Aristotle can’t simply average the 2/11s. Once they have been normalized, we (Thalia, Pascal,  Fermat and others) see 36 cases, of which 6 have coordinates adding to 7.

  10. The use of language is tricky here. The OP begins by saying “one of the dice is a 4”. We understand that to mean “at least one of the dice is a 4”, leading to the answer 2/11.

    But later, in Bonanova’s response, a paragraph begins “one of his dice has a particular value”, and continues as if we are speaking of one particular die.

    Let’s rephrase the original discussion in terms of two distinct dice: a red die and a white die.

    Now Aristotle rolls the dice, and Plato says “the red one is a 4”. What is the probability of a seven total? 1/6. 

    So if we are speaking of one specific die, the answer does not change : 1/6, and Thalia’s answer is right. 

    The phrase “in each particular case he [Aristotle] reasons the probability of seven changes to a new value” shows that Aristotle has been swindled.

    Thank you, Bonanova, this led me to think about the linguistic difficulties faced by Pascal and Fermat, as they initiated the study of probability.

  11. I agree, it is Impossible

     

    E-E can only be 0 or, if there was a borrow from N-S, E-E would be 9

    So N is either 0 or 9
    Now look at the tens position.
    If N =0, there was no borrow, but that requires N>S.
    If N=9, there was a borrow, which means S>N.

  12. is Older = Bill? “Yesterday was Sunday” cannot be true, because Bill lies on Monday, and any other day it would be a lie. And that implies that he lied when he said he’s Bill.

    So Older = John, and he lied when he identified as Bill, so today is either Tuesday, Thursday, or Saturday. Younger (Bill) says tomorrow is Friday. Then Younger (Bill) says he always tells the truth on Wednesday, which is a lie. 

    As they both are lying, today is Tuesday, because it’s the only day they both lie.

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