Hi all.

Can you guys help me in this: give me instances of unique games that test logic, mathematical and spatial abilities please. Games can be cooperative or non-cooperative.
I find it interesting to experiment and to discover the mechanism that makes these games tick :)

Thanks all.

Go.
Hex.
Nim (ok, that's a simple one).
Othello.

Oh my god, I typed in "Othello" in the google image search engine and I ended up getting pornographic materials :(

Ok Nim and Hex will do. Thanks for your help :)

Ok Here is one from MENSA. good luck!


@@@@@@@@@@@@@@@@@@@@?????

The @ means nothing just account for the colors. The question is can you
complete this series of colors? If so, or if not tell me why.
Good Luck
Amy

well if you're going with the light/dark theme I'd guess a light blue could go next

but not a clue if there's more to it such as none of these fit because the specific colors jump all over the rainbow spectrum, not following nature's pattern as I know it would be with these colors =
green yellow pink red

ROY G BIV
Red, Orange, Yellow, Green, Blue, Indigo and Violet

Easy ... this sequence cant be completed ... there exists no color that has a name with just 2 letters

Perhaps you can help me with this riddle that my friend posed to me. I can only solve the first two parts of the riddle but I can't solve the last part. Perhaps you can help me confirm the validity of my first answers to the first two questions and help me with the last? :)

It goes something like this: How many bounded volumes can you get if
a) 2 cubes intersect - my answer: 13 (very sure its correct)
b) 3 cubes intersect - my answer: 67 (im very sure that my methology of getting this answer is correct, but I think the answer may be plus minus 2 since I probably miscount)
c) 4 cubes intersect - ok I have no clue to this one ... good luck if you wish to attempt it :)
in 3 dimensional space.
Let me elucidate a point by what my friend meant by "bounded volumes"
- if 2 circles intersect in 2 dimensional space, you get 3 bounded areas at most.
- if 3 spheres intersect in 3 dimensional space, you get 7 bounded volumes at most.
Hope that clarifies the point.

Please help me solve part c ... good luck :D

Since I have NLD and am extremely bad at visualizing, this is exactly the wrong type of problem for me to solve... :D However, I can tell you, the solution should be forthcoming if you use proof by finite induction. (Given some time, I may be able to solve it for you, too. Don't hold your breath, though. :D )

Hmmm part (b) has proven to be more tenacious and much thornier than I had thought.
Although I'm quite sure that my methodology of solving (b) is correct, but my initial estimate of 67 clearly has overshot. Without the benefit of pen and paper, I believe that the answer to (b) should be around [(1)+(4+1)x4+(6+1)x4] = 47. In order to confirm my answer, I have to draw it out on paper.

Ok I got it! I can't possible be wrong about (b)!
The answer, I'm 100% sure, that it is (1)+(12)+(6)+(24)+(24)= 67!
LoL ... looks like I was right all along. I am willing to BET MY HEAD ON THIS ANSWER: If this is wrong, I will chop my head off!
Good now I can go on to (c).

LoL exeter, if you wish to try question (b) ... BE warned - This question will surely kill you if you do not have a universalistic aka absolute way of accounting for each of the "bounded volumes" individually. That's the best clue I can give you to solve this problem :)

Meanwhile my guess for (c) is 173.

Hey this is fun, can any of you guys post more of these questions here for me to solve?

Hey perhaps someone should start a thread to allow everyone to solve puzzles.

Hey I'm with you buddy! Let's make this thread our puzzle solving thread! Can we be buddies by the way? I can't believe that I finally found someone who clicks at the same pace as I do! I'm only 12, so forgive me for my childishness.
I got the answer for the first one as 13 too and I am sure that it is correct. However I can't figure out the second or third one. Give me some time to solve it please. If I can't solve it will you please show me how?
Oh and as for the color riddle by charlie. I was tricked at first, but I realized that the number of letters decreased for each color, hence the next in the sequence should be a color with just 2 letters, I can't think of a color. Perhaps Link Removed may help. Happy searching!
Anyway, here's a puzzle I came up by myself. Can you guys solve this?
Using only a balance and 11 balls of similar weight and another counterfeit ball that is either lighter or heavier, find out which is the counterfeit one in just 4 weighings, given that every ball looks the same as each other? If this is easy, what about 2 balls that are either heavier or lighter, given that when one is heavier, the other is just as heavy and if one is lighter the other is just as light. What about 3 balls and 4 balls? Good luck!

It's too difficult! I can't get 67 as the answer for part 2! I can get at most 21, being a verifiable answer, to achieve any higher, the cubes would have to intersect in 3D. Yes, for part 1, if they intersect in 3D, you get 13. To get any higher, one would have to make sure each cube intersects the other 2 cubes to give the maximum no. of bounded volumes individually, that is 13. I can get a very faint outline of how this will look like. However, it would be a mental tumult to visualize it, or even draw it out, due to the ambiguity in the mutual intersection of the cubes. I will think about it again. Thanks for your puzzle! Try mine! I'm sure it's no problem for you. No way can a 12 year old gifted child outwit a gifted adult right?

NO...
I worked one entire night on part 2 and I simply can't get the answer! It's too difficult for my puny brain!

After having worked on the problem for 2 hours, I finally got 67 as an answer. Wow you got the answer without the benefit of pen and paper? I don't have the memory capacity to achieve that feat within such a short amount of time.

I see how you got 173 as one of the answers of part 3. It's not easy to resolve part 3 - there are no distinct algorithm which can be used to solve that problem, one has to inevitably depend on heuristics to solve the problem, which is a mental tumult for most people. I won't bother to resolve that, I'm sure I will blow my brain up if I were to spend another hour on this.
By the way, as for the color riddle, I couldn't find a color with just 2 letters. Guess we can't complete the sequence.
Oh yes, as for posting mathematical puzzles here, a word of caution - please try not to post probability questions that deal with infinity - the classical definition of probability will be detered by Bertrand's paradox, uncertainty within statistical probability that limits to infinity.

Oh dear, Avistar sure is smart - he not only solved the riddles that I posted here almost instantly, he also extrapolated my riddle and solved what he had extrapolated in about just 10 seconds. Took me like 20 minutes just to solve all those questions. He suggested that there was no paradox in what I brought out as Bertrand's paradox and helped me to clarify certain uncertainties in the original problem. Well, he told me yesterday that may visit this forum as a guest, not as a member, to help me in some of my riddles! Well, time to bite then :) - will think up of more puzzles and post it here for him to see. We contact each other via email now.

Here's a puzzle I just thought of.

1) You have a box of dimension 10 x 10 x 10.
2) You have several objects:
a) five 1 x 1 x 5 cuboids
b) two 3 x 3 x 3 cubes
c) ten 1 x 1 x 1 cubes
3) How many unique permutations can you slot these objects into the big box, given all of them must fit inside?
4) How many distinct combinations (if you rotate) can you slot these objects into the big box, given all of them must fit inside?
Good luck! I have not solved this yet, I will be sure to give it a look later!