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bonanova

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Posts posted by bonanova

  1. How about ...

    21. Doors

    22. Cars

    23. Led Zep

    24. Kiss [the um, interesting couple under the traffic light]

    25. Madonna [pic in store window - left?]

    26. Red hot chili peppers

    27. Twisted sister - the contortionists in the middle.

    28. Eagles?

  2. All of them. There is only one, which is 2. All other even numbers are divisible by 2 and therefore not prime. The question doesn't specify that the solution must be evenly divisible by 5, therefore ALL numbers are divisible by five, including 2.

    Hmmmm.. well, ok.

    But I've edited the question now, hopefully asking what I had intended to ask.

    See if your answer changes ...

  3. I think this answer

    Yep. I presume you mean 1 mile north of south pole?
    doesn't work: when you get to the South pole, how do you run West?

    But this answer:

    There's an infinite number of circles around the South Pole where he could have started.
    does: for example, any point on the circle (1 + 1/2pi) miles from the South Pole.

    After going South 1 mile, you're (1/2pi) miles from the Pole,

    which allows you to run West 1 mile [1 lap of a 1-mile circumference circle]

    and be able to go a mile North to the starting point.

    As Martini noted, there is an infinite number of starting distances:

    1 + 1/2Npi miles North of the South pole where N is any positive integer.

    N is then the number of circular laps in your westerly mile.

    e.g. N=5280 - you'd run 5280 laps around a 1-foot circumference circle.

    Here's a counter question - why can't N be negative?

    i.e. start closer than a mile - you could still do N laps

    • Upvote 1
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  4. Choose the best answer and say why.

    Which of the even prime numbers [excluding 2] are [evenly] divisible by 5?

    [1] all

    [2] some

    [3] none

    [4] the question has no defensible answer.

    p.s. Martini: I assert this is a logical not mathematical question.

    Edited for "clarity". [meaning that I originally screwed up what I wanted to ask .. ]

    I do not choose [4].

  5. If I understand the "parity" constraint, the idea is to keep neighbors from being nearest or next-nearest in value.

    Neighboring numbers must differ by 3 or more in value.

    I broke the problem down this way.

    First note that 1 and 14 have 11 allowable neighboring numbers; 2 and 13 have 10; the others, 3 thru 12, have 9.

    The 2 most central squares of the grid have 8 neighbors, and their outside neighbors have 6 neighbors.

    The others have fewer: 3, 4 or 5.

    So I start by placing 1-14 in the middle, and 13-2 as their outside neighbors.

    The numbers with the most flexibility are placed in the most demanding squares.

    So we start with this for the middle row:

    x - 13 - 1 - 14 - 2 - x

    Then I try to keep symmetry by placing the remaining numbers in pairs:

    3-12, 4-11, 5-10, 6-9 and 7-8. Start with 3-12 in the outside squares:

    3 - 13 - 1 - 14 - 2 - 12

    and keep going, making sure the lines connecting the added pair pass thru the middle of the grid:

    --------------11

    3 - 13 - 1 - 14 - 2 - 12

    ---------4

    ----10-------11

    3 - 13 - 1 - 14 - 2 - 12

    ---------4--------5

    ----10--6---11

    3 - 13 - 1 - 14 - 2 - 12

    ---------4----9---5

    ----10---6--11---8

    3 - 13 - 1 - 14 - 2 - 12

    -----7---4----9---5

  6. How about ...

    5 4 3 2 8 - can't show the tile boundaries easily, but it should be obvious.

    4 6 7 1 9

    3 7 0 4 2

    2 1 4 1 6

    8 9 2 6 7

    Here's how:

    If a number appears an odd number of times, it has to be on the diagonal.

    Five numbers do that: 0[1], 1[3], 5[1], 6[3], 7[3]

    Upper left UL corner has to be top or left number on a tile.

    Five possibilities: 1[on 3 tiles] 5[on 1 tile] 7[on 1 tile]

    Lower right LR corner has to be bottom or right number on a tile.

    Three possibilities: 6[on 1 tile] 7[on 2 tiles]

    Start eliminating the possibilities. In the order mentioned above,

    Start with UL corner:

    [1] 1-6 tile.

    Requires a 6 beneath the 1.

    No 6 appears on the top or left of a tile. Impossible.

    [2] 1-9 tile. Requires a 9 to right of the 1.

    No 9 appears on top or left of a tile. Impossible.

    [3] 1-4 tile.

    Requires a 4 to right of the 1. Place the 4-6-7 tile there.

    That puts a 6 on the diagonal. OK.

    Also requires a 7 right of the 6. Place the 7-0-4 tile there.

    That puts a 0 on the diagonal. OK.

    Bottom right number must now be 5 or 7. No 5 on bottom or right of a tile, so place 2-6-7 tile there.

    Requires a 5 above the 6.

    No 5 appears on bottom or left of a tile. Impossible.

    [4] 5-4-3 tile.

    Requires a 4 right of the 5. Place the 4-6-7 tile there.

    That creates the diagonal-6. Now the only possible BR number is 7. Place the 2-6-7 tile there.

    That requires a 6 above the 7. Place the 1-6 tile there.

    A 2 must be above the 6. Place the 2-9 tile there.

    A 9 must be left of the bottom-row 2. Place the 1-9 tile there.

    The remaining diagonal number is 0. Place the 7-0-4 tile there.

    4 must be right of the 0. Place the 1-4 tile there.

    Finally place the 2-8 tile and the 3-2-8 tile in their slots.

    Check for symmetry.

    Done.

    [5] 7-0-4-tile.

    Requires a 0 right of the 7.

    No 0 appears on top or left of a tile. Impossible.

    Start with LR Corner

    [6] 1-6 tile.

    Requires a 1 above the 6.

    No 1 appears on bottom or right of a tile. Impossible.

    [7] 4-6-7 tile.

    Requires a 6 to left of 7. Place the 1-6 tile there.

    That places a 1 where a 4 has to be. Impossible.

    [8] 2-6-7 tile.

    Hmm... looks exactly like [4] above. It is. Done.

  7. It was quiet at Morty's last night ...

    Until Matt the Mathematician proposed a game.

    Alex and Davey would both put their billfolds on the table,

    and Matt would count the money in each.

    Whoever had more would forfeit his money to the other player.

    "Think it over," Matt said, and he ordered another frosty one.

    Alex thought to himself,

    "There's no reason to believe I have more money than Davey

    has, or less, for that matter; so my chances of winning are 50%.

    Now if I lose, I'll lose whatever's in my wallet.

    But if I win, it'll be because Davey has more money than I.

    So on a tossup bet, my winnings are more than my losses."

    "I'm in," he finally announced.

    And Davey thought the same.

    He figured his chances of winning were as good as Alex's,

    and he'd either lose the amount in his wallet or win

    an amount greater than that.

    "You're on," said Davey with a smile.

    If the boys were thinking clearly, how could the game favor both of them?

    Or if not, where is the flaw in their reasoning?

  8. 1,177,777

    one million, one hundred seventy seven thousand, seven hundred seventy seven = 23 syllables

    One-mill-ion one-hund-red sev-en-ty sev-en thou-sand sev-en hund-red sev-en-ty sev-en = I count 22.

    But I realize I am wrong with 7,777,771. It should be 1,777,777.

    One-mill-ion sev-en hund-red sev-en-ty sev-en thou-sand sev-en hund-red sev-en-ty sev-en = 23.

    Bravo, Writersblock.

    1,777,777 is the smallest number not specifiable using fewer than twenty-three syllables.

    At least, no one has come up with a smaller number. So let's say it is.

    You get the prize.

    O wait. This is supposed to be a paradox.

    ummm, just for the heck of it, count the syllables in red, above.

    If the red words specified your answer, then ....

  9. For any non-zero, finite number, say 13, for sake of discussion, then usually:

    [1] 0/13 = 0

    [2] 13/0 = infinity

    [3] 0/0 = undefined [that is, could be 242, for example]

    [4] infinity/infinity = undefined [could be 0.327, for example]

    [5] 0 x infinity = undefined [could be 69, for example]

    But it turns out that both zero and infinity have different flavors, called orders.

    There are 1st-order zeros, and 1st-order infinities, as well as 2nd, 3rd-order, etc. of both.

    In these cases, the higher order predominates:

    [3a] 0[2]/0[1] = 0; 0[1]/0[2] = infinity

    [4a] infinity[2]/infinity[1] = infinity; infinity[1]/infinity[2] = 0.

    [5a] 0[2] x infinity[1] = 0; 0[1] x infinity[2] = infinity.

    where 0[n] means n-th order 0 and infinity[p] means p-th order infinity.

    You can see what that means by changing one of these zeros a wee bit, then inching your way back.

    Let x be a small number - say 1/100.

    [6] x/x=1. no problem, since x isn't zero. Let x inch its way to zero, the ratio stays at 1.

    [7] 2x/x = 2. no problem, since x isn't zero [yet]. Let x -> 0, the ratio stays 2.

    That's why 0/0 is usually thought of as undefined: you could have used any number instead of 2.

    But look at

    [8] x^2/x = x. now the ratio is x. Let x ->0, and the ratio [always = x] ->0 as well.

    What happened? both numerator and denominator -> 0, so the ratio goes to 0/0.

    How come it's not undefined? How come in this case we know it's zero?

    The reason is x^2 becomes a 2nd-order zero: 0^2 so to speak; it's stronger than just 0.

    Same thing about the infinity you get by dividing by x^2 instead of by x as x->0.

    There you get a stronger infinity - a 2nd-order infinity - infinity^2 so to speak.

    There is a rule called L'Hopital's rule that gives you these answers for [6] - [8].

    Read more about that here: http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

    Basically you step away from the divide-by-zero situation and evaluate the division using a legal small number, then see what happens when it approaches zero, and that gives you a defined answer, for that particular case.

    L'Hopital's rule can make some paradoxical problems intuitive.

    Absent some expression like [2x/x] ->2, or [x^2/x] ->0, or [x/x^2] -> infinity, you're just left with undefined and you don't try to prove too much with expressions that contain infinities in the numerator or zeros in the denominator.

  10. If they are bald they have no hair to be judged by. They can not have the same amount of hair on their head
    Interesting.

    0 does not equal 0.

    But let's keep this a secret, OK?

    Word gets out, imagine the recall of all those text books.

    Would be devastating.

  11. The inscriptions on Gold and Silver agree, so they must have the same truth value.

    By conditions of the problem, Lead has the opposite truth value.

    Gold and Silver False means

    [1] Ring is in Silver

    [2] Lead is true -> Ring is in Lead.

    Contradiction

    Gold and Silver True means

    [1] Ring is in Gold or Lead

    [2] Lead is false -> Ring is not in Lead.

    Ring is in gold box.

  12. That's it.

    Here's my method: [care to share yours?]

    The number has to end in 9.

    Looked brute force for small numbers.

    59 and 119 were promising, but no cigar.

    Then looked for agreement among

    39 + multiples of 40,

    69 + multiples of 70 and

    89 + multiples of 90

    Smallest one was 2519.

    Still think of this as kind of brute force.

    Maybe there is no elegant solution.

  13. Considering only positive integers, 7 is unambiguously specified by the phrase the smallest number not specifiable using fewer than two syllables.

    What is the smallest number not specifiable using fewer than twenty-three syllables?


    How many syllables does the smallest number not specifiable using fewer than twenty-three syllables have?
  14. I just found a number with an interesting property:

    When I divide it by 2, the remainder is 1.

    When I divide it by 3, the remainder is 2.

    When I divide it by 4, the remainder is 3.

    When I divide it by 5, the remainder is 4.

    When I divide it by 6, the remainder is 5.

    When I divide it by 7, the remainder is 6.

    When I divide it by 8, the remainder is 7.

    When I divide it by 9, the remainder is 8.

    When I divide it by 10, the remainder is 9.

    It's not a small number, but it's not really big, either.

    When I looked for a smaller number with this property I couldn't find one.

    Can you find it?

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