BrainDen.com - Brain Teasers

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## Popular Content

Showing content with the highest reputation since 01/13/16 in all areas

1. 1 point

## Connect the Dots

and a Spoiler for some additional thoughts:
2. 1 point

## 3D Connect the Dots

just guesing without doing calculation.
3. 1 point

## 3D Connect the Dots

Connecting the 8 corks with web with least linear meter requires connecting with the web along the sides of the cube (no diagonals) with length 1m, and uses at least 7 segments to connect 8 corks. so the total length is 7*1m = 7 meters. and the shape could vary, as long as you pass the web only along the sides of the cube.
4. 1 point

## Connect the Dots

I finally found a way to make the road system less than 7.5 miles.
5. 1 point

## logic sequences

Hi wiseabel, and welcome to the Den. So clearly there are several ways in which these numbers are similar and, equally clearly, we're to look for something beyond the fact they are integers. Three are odd, and a different group of three are prime. Three of the four contiguous pairs are descending. Only one of the digits (1) repeats among the numbers. At first glance, I don't see a common similarity, but I'll give it some more thought perhaps in a later post. Thanks for submitting a puzzle!
6. 1 point

## 3 remarkable numbers

I'll bet this is not the answer you're looking for, but it does qualify as a remarkable matchup:
7. 1 point

## unknown riddle

to life or not to life this code itself is a toast a jewish custom and saying when it is time to boast i would raise my glass in a gathering over roast a wedding or maybe just to avoid becoming a ghost i toast to the counter sign and toast to life but the word has just slipped my mind remind me? hey guys, i need some help for this riddle, since i cant get any out of it.
8. 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.
9. 1 point

10. 1 point

## rang the bell.

A man enter a room. In the room there is a 2 digits number on the wall. His friend outside the room do not know the number. But just by ring the bell once, his friend know the number. how could this happen ?
11. 1 point

## Rights Of Passage

and that led me to this..
12. 1 point

or
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15. 1 point

## Math Lover?

"You have an OCD of things in even numbers" "No I have not... . . . . No I have not."
16. 1 point

9^(9-9)=1
17. 1 point

## Lychrel

Also: http://projecteuler.net/problem=55
18. 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; } } }
19. 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?
20. 1 point

## 1000 people in a circle

Person 1 is left behind there would be no one to kill him
21. 1 point

Yes, that makes sense. It's like the guy who was prophesied to be killed by his son, so he has all his sons put to death, but one escapes death, grows up not knowing his father, and winds up killing him because of it. Destiny & free will are not mutually exclusive. That upsets people.
22. 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.
23. 1 point