Jump to content
BrainDen.com - Brain Teasers


  • Content count

  • Joined

  • Last visited

  • Days Won


Posts posted by Thalia

  1. I got the outfield and 2B without assuming children have married fathers. The rest seem to depend on that assumption. I've seen a couple sites post this with those assumptions stated with the original problem. I found the sports illustrated version but it sounds like the writer had gotten it elsewhere as he mentioned he hadn't solved it yet. So that is not the original source.

    It looks like the only purpose of mentioning the residence is to establish that a given player does not play a given position. Sounds more interesting than just saying "Adams is not the catcher." or "Jones is not the third baseman."

  2. Thanks for the clarification. After some googling, I see apartment house is another term for apartment building. . . So if you live in an apartment, you live in a "house". Haven't done the puzzle yet so not sure how relevant that is but it would explain the perceived discrepancy. It was noted that this was published in 1959. I'm sure reality hasn't changed since then in terms of children and bachelors but I'm wondering if there were some different assumptions back then about recognizing children outside of marriage. I get the feeling that unless you're a politician, it might not be quite as big of a deal for a bachelor (or bachelorette) to have a child now as it was in the 50s. I don't have any evidence of that though.

    How does recognizing divorce contradict engaged men being bachelors? Google shows a bachelor as being a man who isn't and has never been married. So an engaged man would be a bachelor because he's not actually married, a divorced man would not be a bachelor because he used to be married, but as written, it doesn't matter anyway because Green is still married until the suit is settled.

  3. 3 hours ago, Shanna said:

    I found this way late in the game, but this puzzle makes no sense.  If we are to assume that some people are married or single, then are we also to assume that the word "house" means the same thing as the word "apartment"?  I found this puzzle a few years ago and forgot about it until I found it today.  I assumed that Adams cannot be the third baseman because he lives in a house while the 3B lives in an apartment.  I got online to search for the answer because, of course, the puzzle is not solvable.  It seems that Sports Illustrated first published it in 1959 and reposted it online.  It then apparently went viral.  One thing I can say is that any site that has this puzzle posted has people discussing it and everyone agreeing with each other, the same as in this forum, that Adams is the 3B.  Not a single person has questioned the difference in residence type.  This is the worst puzzle for someone like me because little clues like that are usually given to help answer the puzzle.  

    "3. The third baseman lives across the corridor from Jones in the same apartment house."

    Normally, I would consider an apartment and a house to be different. But clue number 3 seems to indicate that they are interchangeable.

  4. Spoiler

    A few thoughts:

    It would be better to start the ball swinging at the beginning to give yourself the most amount of time.

    You could also hook the string with the pipe as the balls are swinging so you can catch both balls on the inside of the pipes. At that point, you could attempt to flip the balls up like that toy where the ball on a string is attached to a cup.

    He said you can't touch it with your hands. He didn't say anything about feet so I suppose if you're extremely flexible. . .

    Just because it's possible doesn't mean it's easy even if you have a good strategy. . .


  5. 15 hours ago, Pickett said:

    Haha, that's kind of fun to wrap your head around...

      Hide contents

    But really, it should be fairly simple, I think:

    Each option has a 25% chance of being randomly picked. That is the correct answer: 25%.

    BUT there are two options where "25%" is the answer. This means that the answer that you pick has a 50% chance of being "25%", which is STILL the correct answer regardless of the probability of picking an answer that HAS that value. So the probability that you are correct is 50% which is answer 2...



    The probability that you choose the correct value of 25% is 50% but since 50% is not 25%, you'd be part of the 50% that guessed wrong?


  6. On 3/15/2018 at 5:20 PM, flamebirde said:

    Incidentally, the title was a hint too: plain in color, and also plain -->plane, since a piece of paper is flat.

    Lol. That's what lead to one of the other guesses. Plain to see--> crystal clear.

    Crystal would be fancier drinking cups though. And I couldn't quite call it white. 

  7. Got the same letter and number equivalents as Pickett. I'm reading it as 

    ONE 1(ONE) + 2(ONE) + 4(ONE) + 17 = ONE (1+2+4)(ONE) +17 = ONE 7(ONE) +17 = ONE DONE MD? No clue what that means though...

  8. Well, there goes your curve estimation. I don't remember the situation but there were certain combinations that I could calculate with some multiplication. I think one of them was a CE combination. But of course the other combinations didn't feel like cooperating. I'll try to work on this over the weekend. 

    Agree on your pink combinations except for XMEE. Should be the same count as EEM, right?

  9. I did some quick counting for the easier ones. I've noticed that at least up to this point, the number of combinations of C, E, M, and X is (# cubelets removed+1)^2. There are 25 combinations for 4 cubelets removed. The ones including X would be the same as the counts for 3 cubelets without X. I did not count the combinations that have 2 or more E's or CCEM, CCMM, AND CEMM. I got 191 for the rest. I'm guessing the ones I didn't count are going to be pretty high so maybe around 500?

  10. 3 hours ago, rocdocmac said:

    @Thalia ...

      Hide contents

    I actually noticed something in this line yesterday!

    T26<>1: If 26 cubelets are removed, one would be left with either X, M (= F), C or E, i.e. T26 = 4 (not "1"!).

    We now know that T3 = 139.

    Thus T0, T1, T2, T3, ... = 1, 4, 22, 139, ...

    The sequence should therefore end with

    ... T24, T25, T26, T27 = ... 139, 22, 4, 0

    Maximum values would be for T13 = T14


    When you remove a cubelet, the kind you remove matters because it changes the shape. When you take away 26, it doesn't matter which one is left. There may be 4 different labels but in the end, when it comes to the remaining shape, all you see is a cube.

    Another thing. For T27, there is one possibility. But given the phrasing of the original question, does nothing count as a shape?