BrainDen.com - Brain Teasers

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## Popular Content

Showing content with the highest reputation since 01/03/17 in all areas

1. 1 point

## Senseless promotions? Or, the wisdom of the tea people

EDIT: For clarity, I use "number" to reference an individual digit. I use "sequence" to reference the string of numbers. EDIT2: Added spoiler tag.
2. 1 point

## Video Game Logic

Perhaps your question may hinge on the source of variability. Pistol has 5-7. In any skirmish, is the choice between 5, 6, 7 made by the program randomly? Or does it depend upon distance to target? Or visibility ( like heavy, light or no fog)? And, as Pickett points out, damage radius is a key parameter of the Dupuy lethality index.
3. 1 point

## Video Game Logic

Another factor to possibly consider...
4. 1 point

Depends...
5. 1 point

## Coin hunt

Fuzzy thoughts:
6. 1 point

## Worth the Weight

You are shown a pile of dimes all of which have one of two distinct weights differing by a small amount not detectable by feel. Forty eight dimes are separated from this pile and you are told of these forty eight, light ones are a dime a dozen (literally - i.e. 44 heavy dimes and 4 light dimes). Using a balance scale twice, find seven heavy dimes. EDIT: for clarification
7. 1 point

## Worth the Weight

4 light ones/48
8. 1 point

## Thank goodness, no remainder is chasing me

Ten years ago I called attention to a number that when divided by a single integer p it left a remainder of p-1. (Help, a remainder is chasing me) Here is a chance to construct a nine-digit number, a permutation of { 1 2 3 4 5 6 7 8 9 } that has no remainders, sort of. The task is to permute { 1 2 3 4 5 6 7 8 9 } to create a number whose first n digits is a multiple of n for any single-digit n. For example, consider 123654987. Its first 2 digits (12) are divisible by 2. It's first 5 digits (12365) are divisible by 5. However this is not a solution, since 1236549 is not a multiple of 7.
9. 1 point

## Chess Board Puzzle

Use the twelve pieces on the left to create a chess board, The pieces may be flipped!
10. 1 point

11. 1 point

## Westworld Mafia

Does the Indy trap appear even if it doesn't save anyone from actual death? Can they act on themselves? Do the goodie save and the Indy trap appear identically in the night post?
12. 1 point

## Westworld Mafia Signups

Signups: 1. Gavinksong2. Flamebirde3. Nana774. plasmid5. phil18826. bonanova 7. aura8.9. 10.11.12 I've just moved, to Ohio, from NY and drowning in the details of it. Reluctant to pledging my attention here. But, I'll give it a shot ... Should be able to check in at least daily.
13. 1 point

## Probability of selecting two blue discs back to back

If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, P(BB) = (15/21)×(14/20) = 1/2. The next such arrangement, for which there is exactly 50% chance of taking two blue discs at random, is a box containing eighty-five blue discs and thirty-five red discs. By finding the first arrangement to contain over 1012 = 1,000,000,000,000 discs in total, determine the number of blue discs that the box would contain.
14. 1 point

## Two people cross a river

Conjoined twins?
15. 1 point

## A bug problem

Sorry, something got lost in copying and pasting. The site does not like the formatting used in word, so I am attaching a picture of the solution ... apologies in advance.
16. 1 point

## Lychrel

import java.util.*; public class lychrel { public static long palin(long n) { long m,p=0; while(n>0) { m=n%10; p=(p*10)+m; n=n/10; } return p; } public static void main(String ar[]) { long a,l,i,j=0,t,it=0; Scanner v=new Scanner(System.in); a=v.nextInt(); l=a+palin(a); t=palin(l); for(i=1;i<=10;i++) { it++; if(l==t) break; else { l=l+palin(l); t=palin(l); } if(i==4) j=l; } if(l!=t) { System.out.println(a+" is a Lychrel Number"); System.out.println("5th iteration of number "+a+" is "+j); } else System.out.println(it); } }
17. 1 point

18. 1 point

## Mouse & Cheese Cube

Hopefully this one has not appeared before... Suppose 27 identical cubical chunks of cheese are piled together to form a cubical stack, as illustrated below. What is the maximum number of these cheese chunks through which a mouse of negligible size could munch before exiting the stack, assuming that the mouse always travels along the grid of 27 straight lines that pass through the centers of the chunks parallel or perpendicular to their sides, always makes a 90 degree turn at the center of each chunk it enters, and never enters any chunk more than once?
19. 1 point

## I'm not a Trump supporter

We're being intruded by migrants secluded Who stowaway past our frontier They're costing us green and will linger unseen Unless I should get into gear Their nature I'll dredge as I perch on a ledge And twisting the truth's my endeavor Exposed to the nation, and next: deportation They thought they'd endure thus forever? I promise the answer has nothing to do with politics
20. 1 point

## Forgotten District II Mafia

Hirk had to have been a goodie: baddie wouldn't target himself for a kill, and Indy can't be killed N1.
21. 1 point

## Forgotten District II Mafia

Edit - double post, so I thought I'd take this time to post a dancing monkey gif. Enjoy.
22. 1 point

## Forgotten District II Mafia

Can I be the one to own you Molly? Or be owned by you? Please
23. 1 point

Player A has one more coin than player B. Both players throw all of their coins simultaneously and observe the number that come up heads. Assuming all the coins are fair, what is the probability that A obtains more heads than B?
24. 1 point

## Hard riddle

(BR / BL +UR) UL Thus... (8/4 +2) 3= 12 (2/1 + 6) 4= 32 (9/3 +4) 2= 14 and for the last one.. (6/2 +3) 5 = 30
25. 1 point

## Signups for Forgotten District II Mafia

On my phone, I see a link on each post that says "quote". I can sort of copy and paste the roster on phone but as I found out last game, the cursor had an unfortunate habit of moving... Spoilers and colors on phone have to be done with the [ ] codes. Spoiler: {spoiler}your text here{/spoiler} with [ ] instead of { }. No spoiler titles. Colors: {color=red/orange/whatever}text here{/color} also with [ ] instead of { }.
26. 1 point

## Signups for Forgotten District II Mafia

@MissKitten Tagging does not work on mobile and quoting is hard on mobile as well. But other than that, on a PC, one can copy-paste colors and not bother with quoting the roster altogether. Which is cool. At least from a host's PoV.
27. 1 point

@SakuraChan
28. 1 point

## The Coup of Rhotus Mafia Signups

I think it's fair to say that the Coup of Rhotus is effectively dead for now; we're having enough trouble getting signups for a ten player game as is. Maybe once we build the player base a bit we can try to revive it, but for now I think it's rather unlikely. On the other hand, Forgotten Districts is in signups, and that's relatively close to getting off the ground!
29. 1 point

## Signups for Forgotten District II Mafia

Roster: 1. maurice 2. Flamebirde 3. Molly Mae 4. 5. 6. Hirkala 7. bonanova 8. 9. 10. Roster: 1. maurice 2. Flamebirde 3. Molly Mae 4. 5. 6. Hirkala 7. bonanova 8. 9. 10. Mafia!!! Been too long! I don't know if I was supposed to leave it doubled, but I did. Hi, everyone!
30. 1 point

31. 1 point

32. 1 point

## UN Mafia III

If it makes you feel any better, I had you and Molly as Germany and Russia before the mass outing. Thought bonanova sort of claimed Australia earlier but figured if the US killed him I'd be better off whether he was Australia or China. He seemed like the biggest threat to me with his day actions. If maurice had removed a vote besides flamebirde's, it would have been a tie and I wouldn't have even got the chance to use my RID kill. When I was first reading through, I didn't realize you were expecting me to kill Molly. Got a bit of a laugh from that. So the biggest thing the mass outing accomplished was it made sure you got China. If you had lynched fb, I would have RID killed maurice which would have been blocking. I did get really lucky with bonanova getting the UK lynched D1 so thanks for that. On another note, I came extremely close to not acting at all N2 to try to frame plasmid as Sweden. Might have given mo's claim more credibility with neither Sweden nor China showing... So now that I know asking won't get me killed, what's a PGP public key block? And where were those soft claims?
33. 1 point

## UN Mafia III

Hey guys, sorry I'm here. Didn't realize I could talk when I was trapped and then I'm having issues accessing the site on my phone. I'm on through Firefox now and it works better than Chrome I guess. I will catch up at lunch and see what I can contribute, hopefully before the night ends in case I was mercy/mod killed.
34. 1 point

## UN Mafia III

@Thalia, good points. I've played only a few Mafia games. I wish I had accepted Unreality's invite to play, back in the day, and learned it better. Coupled with a hope for a few more games to follow this one. My comfort zone is math and logic, where all the needed info is provided, just needing insight and clear thought. This is more fun, actually, and with experience I might get better. Or, I'm a very slippery guy and everything I just said is garbage. . When does the night end?
35. 1 point

## UN Mafia III

@Thalia Understand, (with apologies to the players.) Hanlon's Razor. Seriously, what do we know about maurice?
36. 1 point

## UN Mafia III

Taps watch. Oops. RIP MikeD. Would love to hear from maurice.
37. 1 point

38. 1 point

## Signups for a Mafia game?

I was thinking more like 7 = 1 baddie vs 5 goodies vs 1 indy, hosted 2 Mafias the same size recently so I'm pretty sure I can balance it in a classical setting. If no one signs up till Friday night, I'll trim it, send out the roles and N1 can end Monday. The individual wincons can't be balanced for a trimmed version unfortunately. Anyone leaving the game would make it lopsided and I don't see enough ideas of individual wincons to make a miracle in-game balance for events. But I guess I can run it again in the future as a redux of the OP if there are more people interested during the summers for example. So, all is not lost. I'd rather have you play since you all showed up and signed up
39. 1 point

## Binary lock

Real life: I bought (very cheep) a 256MB extension card for a 80286. There were 8 dip switches and no manual. I do not know how many times each switch can be flipped, in any case, it is better to minimize the manipulation. Look https://en.wikipedia.org/wiki/Gray_code
40. 1 point

## The Donny Hall puzzle

Around World Series time recently, (that's the American baseball championships for our international Denizens) I asked the question, dressed as a probability puzzle, "Given that you could pick one of the best-of-seven games in the series that you could guaranty to win, which game would that be? I even flavored the question with the opinions of experts that the pivotal games are often taken to be game 3 or game 5, as statistically these games correlated well with the eventual winner. And many responded with one of those choices for their answers. But the correct answer of course is game 7. (Or as some said, the Last game.) So I also like the "one with the car" choice.
41. 1 point

## The Donny Hall puzzle

I think phaze is right about the most practical answer. If that fails, I tend to agree with bonanova.
42. 1 point

## Folding paper

Agree. When I hit the send button, I realized my thinking was too simple. But instead of deleting my post (moderator privilege) I left it to take its licks.
43. 1 point

I believe you replace the ? with digits.
44. 1 point

## IDENTIFY THE FAKE COIN

Divide the thirty coins into four groups of 9, 9, 9, and 3. Label the 9 coin piles as 9a, 9b, and 9c. Weighing number one is 9a vs. 9b. If they balance, the odd coin is in either in 9c or in the pile of 3. If so, then weighing Number 2 is 9a vs. 9c. If they balance, then the odd coin is in the pile of 3. Label the three coins the pile of 3 as 3a, 3b, and 3c. Weighing number 3 is 3a vs. 3b. If they balance, the odd coin is 3c. In that case, weighing number four is 3c vs. any other coin to determine if the odd coin is lighter or heavier than the rest. If after weighing No. 3, coin 3a does not balance against coin 3b, then one of them must be the odd coin. Remove the lighter coin, let’s say it’s 3b, and place coin 3c on the scale against the remaining heavier one, 3a. If they balance, then the odd coin is 3b and it is lighter. If they do not balance, then the odd coin is 3a and it is heavier. Now let’s go back and see what happens if the first weighing has a different outcome. Weighing number one was 9a vs. 9b. Let’s say that they do NOT balance. That means that the odd coin is in either 9a or 9b. Note which of the two is heavier. Let’s say 9a is heavier. Weighing number two is 9a vs. 9c. They will either balance or 9a will be heavier. There is no possibility that 9a will be lighter. If 9a and 9c balance that means that the odd coin is in 9b and that it is lighter. If 9a is again heavier, that means that the odd coin is in 9a and it is heavier than the rest. Let’s say it’s in 9a and it’s heavier. Now divide 9a into three groups of three coins each. Label them 3A, 3B, and 3C. Weighing number three is 3A vs 3B. If they balance then the odd (heavier) coin is in 3C. If they do not balance, the odd (heavier) coin is in the heavier of 3A or 3B. Take the group of the three that contains the odd coin and divide it into 3 individual coins: 1a, 1b, and 1 c. The fourth weighing is 1a vs. 1 b. If they balance the odd is is 1c and it is heavier. If they do not balance, then the odd coin is the heavier of 1a and 1b. Using this same methodology, the odd coin and its status may be determined no matter how each of the weighings turn out.
45. 1 point

9^(9-9)=1
46. 1 point

## Lychrel

Also: http://projecteuler.net/problem=55
47. 1 point

## Lychrel

Code: #include <iostream> #define BUF_SIZE 100 using namespace std; void putInArray(int num, int* buf){ for(int i=0;i<BUF_SIZE;i++){ buf[i] = num % 10; num = num / 10; } } bool isPalindrom(int* buf){ int i=0, j=BUF_SIZE-1; while(buf[j]==0 && j>0){ j--; } while(i<j){ if(buf[i] != buf[j]){ return false; } i++; j--; } return true; } void applyIteration(int* buf){ int i=0, j=BUF_SIZE-1; while(buf[j]==0 && j>0){ j--; } while(i<=j){ int temp = buf[i] + buf[j]; buf[i++] = temp; buf[j--] = temp; } for(int i=0;i<BUF_SIZE-1;i++){ buf[i+1] += buf[i] / 10; buf[i] = buf[i] % 10; } } bool isLychrel(int* buf){ for(int i=1;i<50;i++){ applyIteration(buf); if(isPalindrom(buf)){ return false; } } return true; } int main() { int buf[BUF_SIZE]; for(int num=0;num<10000;num++){ putInArray(num,buf); if(isLychrel(buf)){ cout << num << " is Lychrel " << endl; } } }
48. 1 point

## I'm not a mythical titan

Excellent riddle Plasmid. Those types are my favorite. Give yourself a 5 star vote and make it official. BTW . . . what's the "vote" thing for?
49. 1 point

## 1000 people in a circle

Person 1 is left behind there would be no one to kill him
50. 1 point

## Is it possible to give what we don't have?

Depends on what is meant by "have." [1] If have means "own," then yes. I can give someone something that I do not own. e.g. if I stole it. To give something, one only needs the ability to determine who controls it. If I control it, I can pass its control to someone else. So ... [2] If have means "possess the control of" then no. As stated, the paradox arises from the different antecedents of "with sorrow." Sorrow is the consequence of giving, not a possession before the act. But the language permits that interpretation by its form. Cute.
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