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Pickett

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Posts posted by Pickett

  4. Relatively prime

    in New Logic/Math Puzzles

    Posted

    So, if I am correct in assuming "relatively prime" simply means that one number is not evenly divisible by the other, then here are my initial thoughts:

     

    You have a 1/n chance of first drawing a "1"...in which case, you will win automatically (1/1 chance)

    You have a 1/n chance of first drawing a "2"...in which case, you will win if the second number is even OR a factor of 2 (aka: 1)

    You have a 1/n chance of first drawing a "3"...in which case, you will win if the other number is a multiple of 3 OR a factor of 3 (aka: 1)

    You have a 1/n chance of first drawing a "4"...in which case, you will win if the other number is a multiple of 4 OR a factor of 4 (aka: 1 or 2)

    ...

    You have a 1/n chance of drawing "x"...in which case, you will win if the other number is a multiple of x OR a factor of x

     

    So let FACT(x) = number of factors of number x...we can then write our chance of winning to be:

     

    SUM (1/n * (1/x + FACT(X)/n)), x=1..n

    Which clearly just get worse and worse the more balls there are in the bin...in which case, I would only take the bet if there were visibly "few" (less than 20 or so) balls in the bin.

     

    Now, I suppose my above reasoning only holds if we assume all the numbers 1-n are in the bin...if you have completely random/arbitrary numbers...then, my reasoning probably completely falls through. Again, just my first thoughts. I am also curious to see what others come up with.

  5. How Many Fingers Am I Holding Up?

    in New Word Riddles

    Posted

    I just saw this puzzle...it's pretty awesome...

    All you really need to do is notice that the string ends with "=="...which, if you know anything about encodings, is a common ending for Base64 encoded strings.



    So, just plugging the string into a Base64 decoder, you get the result that was flagged as the answer.

    Really neat puzzle...simple if you get that first step...and props on the ASCII art!
    • Upvote 1

  6. Parts of Words

    in New Word Riddles

    Posted

    mic

    roscope

    paradiddle (if you're a drummer, this makes sense...all I could come up with!)

    ???? I've got nothing for "LTIQ"...could go French and say "Celtique", but that's the best I can think of...

    dogwood

    hydrolysis

    horseplay

  10. Covering triangles with triminoes

    in New Logic/Math Puzzles

    Posted

    6: no - after putting a triangle in each corner (which must be done to fill the corners), you must put another triangle with its edge along the edge of the board, which then leaves you with 3 separated hexagons left
    7: no - 28 hexagons on the board is not divisible by 3

    8: no - again, after putting the triangles in the corners, you will end up with 3 hexagons that left over that can't be covered with a triangle

    9: yes

     

    I can give more detail if needed for any of these...but the OP didn't actually ask for "why or why not..." so...

  11. Classic Potato Problem

    in New Logic/Math Puzzles

    Posted · Edited by Pickett

    Problems like this always SEEM so straightforward, but the answers are always a little shocking to see:



    New weight: 50lbs
    Weight of the 1% water lost: 50lbs
    Weight of the 98% water remaining: 49lbs

     

    First let's do the obligitory/obvious initial analysis of the original amounts:

    0.99 = water weight / total weight

    0.99 = water weight / 100

    water weight = 99lbs

     

    So, now we need to find the new weights:

    0.98 = new water weight / new total weight

     

    we know that the new water weight is: 99 - water weight lost

    and we know the new total weight is: 100 - water weight lost

     

    Let x = water weight lost

    This gives us the equation: 

    0.98 = (99 - x) / (100 - x)

    98 - .98x = 99 - x

    0.02x = 1

    x = 50

     

    This means the water weight lost was 50lbs...so using that, we can find the answers above. We can certainly validate this answer by showing that if we have 50lbs of potatoes, and 49lbs of it are water, we indeed have 98% water weight, and both the total weight and water weight decreased by the same amount (50lbs)...or another way of seeing that is that the original weight of the non-water parts of the potatoes were 1 pound (100 - water weight)...since that does not change after the water evaporates overnight, the end result should still have 1 pound of non-water weight...which we do have (50 - 49).

  13. riddles

    in New Word Riddles

    Posted

    Potatoes


    Needles
    Hurricanes/tornadoes/storms
    Photographs
    A blind person/animal
    Dolls

    if you go by just HEARING this riddle (instead of reading it written out), you could say words (like "Mississippi" has 4 "eyes")

    Lots of answers here...

  14. Unreal numbers

    in New Logic/Math Puzzles

    Posted

    423 + 675 = 1098


    425 + 673 = 1098
    429 + 876 = 1305
    432 + 657 = 1089
    437 + 652 = 1089
    439 + 587 = 1026
    489 + 537 = 1026
    539 + 487 = 1026
    589 + 437 = 1026
    623 + 475 = 1098
    625 + 473 = 1098
    629 + 874 = 1503
    632 + 457 = 1089
    637 + 452 = 1089
    724 + 365 = 1089
    725 + 364 = 1089
    742 + 356 = 1098
    746 + 352 = 1098
    749 + 286 = 1035
    749 + 853 = 1602
    769 + 284 = 1053
    789 + 246 = 1035
    789 + 264 = 1053
    829 + 476 = 1305
    829 + 674 = 1503
    849 + 357 = 1206
    • Upvote 1

  16. Days of Dates

    in New Logic/Math Puzzles

    Posted · Edited by Pickett

    The calendar repeats itself AT LEAST every 28 years (could be every 5, 6, 11, or any combination of those...)...so if we find one year that matches the above calendar, we would be able to find infinite many...

     

    With that being said, the NEXT year that fits this pattern will be 2017...and the LAST one that did was 2006 (notice how that is 11 years apart, as mentioned above)...

     

    Here are the years in the 2000s that fit this pattern:

    2006, 2017, 2023, 2034, 2045, 2051, 2062, 2073, 2079, 2090

     

    And here are the years in the 1900s that fit this pattern:

    1995, 1989, 1978, 1967, 1961, 1950, 1939, 1933, 1922, 1911, 1905

     

    ...That is, unless I'm missing something in this puzzle that helps us pin-point an exact year...and that I'm interpreting the puzzle correctly :c)

  17. Special number

    in New Logic/Math Puzzles

    Posted · Edited by Pickett

    6210001000



    I can explain the process that I used in more depth if needed...but it's really not too difficult if you begin by choosing the first number in the randomly generated number and start filling in the inventory number step by step based on that...I started this process with the number starting with 9 and worked my way backwards, but you could do it starting at 1 and work your way up.

    As you do that, you'll find that the generated number can't start with a 9, 8, or 7...you eventually get to numbers that start with 6:
    6 ? ? ? ? ? ? ? ? ?
    ? ? ? ? ? ? 1 ? ? ?

    6 ? ? ? ? ? 1 ? ? ?
    ? 1 ? ? ? ? 1 ? ? ?

    6 1 ? ? ? ? 1 ? ? ?
    ? 2 ? ? ? ? 1 ? ? ?

     

    Have to change our generated number to match the inventory number...

    6 2 ? ? ? ? 1 ? ? ?
    ? 2 1 ? ? ? ? ? ? ?

    6 2 1 ? ? ? 1 ? ? ?
    ? 2 1 ? ? ? 1 ? ? ?

    6 2 1 ? ? ? 1 ? ? ?
    6 2 1 ? ? ? 1 ? ? ?

    6210001000

    • Upvote 1

  21. Decimal Digits

    in New Logic/Math Puzzles

    Posted · Edited by Pickett

    not sure if it's the smallest yet, but it's probably getting pretty close at least:

    .0123456789

    .0123456798

    .0123456879

    .0123456897

    .0123456978

    .0123456987

    .0123457689

    .0142675398

    .0196582374

    .1203456789

    I'm thinking this HAS to be the smallest possible sum...here's why:

    The smallest valid number we can use is .0123456789

    So if we do that 9 times (illegally, of course...but just doing it for the sake of demonstration), the sum would be .1111111101...which would be the absolute minimum value for the 10th number.

    Obviously that isn't a valid number (according to the constraints of the problem), so the closest number we can get to that absolute minimum would be .1203456789

    Which is the answer I have above.

    Now, I there are multiple WAYS to get that sum, but it should be the minimum sum.

  23. Minimizing the number of terms of the square of a polynomial

    in New Logic/Math Puzzles

    Posted · Edited by Pickett

    First off, expanded out:

    P(x) = x8 + 2ax7 + (a2 + 2b)x6 + (2ab + 2c)x5 + (2ac + b2 + 2d)x4 + (2ad + 2bc)x3 + (2bd + c2)x2 + 2cdx + d2

     

    So that means all 9 coefficients are:

    1. (1)
    2. (2a)
    3. (a2 + 2b)
    4. (2ab + 2c)
    5. (2ac + b2 + 2d)
    6. (2ad + 2bc)
    7. (2bd + c2)
    8. (2cd)
    9. (d2)

    Since we know a, b, c, and d are non-zero...that means coefficients 1, 2, 8, and 9 can never be 0. Therefore, the theoretical max would be 5 coefficients equaling zero:

    1. (a2 + 2b) = 0
    2. (2ab + 2c) = 0
    3. (2ac + b2 + 2d) = 0
    4. (2ad + 2bc) = 0
    5. (2bd + c2) = 0

    By equation 1, we can see that b = -(a2/2)
    Then by substitution in equation 2, we can find that c = a3/2
    Then using substitution in equations 3, 4, and 5 we find the values of d...which end up being:
    By equation 3: d = -(5a4/8)
    By equations 4 and 5: d = a4/4

    Since there are no situations where -(5a4/8) = a4/4 are true (except when a = 0, which we know it doesn't), we have to choose the one that gives us more valid solutions, which is that d = a4/4...which means that we can have a maximum of 4 coefficients that equal 0 in this expanded equation (which means the minimum non-zero would be 5):

    • a = x
    • b = -(x2/2)
    • c = x3/2
    • d = x4/4

    We now need to find values of x that give us integer values...so an obvious answer would simply be x=4 and/or x=-4, which gives us these two solutions (resulting in 4 zero coefficients, aka 5 non-zero terms):

    SOLUTION 1:
    a = -4
    b = -8
    c = -32
    d = 64

    SOLUTION 2:

    a = 4
    b = -8
    c = 32
    d = 64

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