Thinking hurts. However, stupidities come to world without birth pains.
Bielaszewski
Best Puzzles
This is collection of the best puzzles selected out of +10K brain teasers. Link to solution is under each puzzle. The answer is usually in spoiler which you can unhide by clicking the "Show" button. Some of them might be quite hard, however, you don't need any deep math knowledge. Just basic logic, creativity and patience is needed.
Here they come - BrainDen Top 10 Puzzles.
Enjoy!
1. Weighing in a Harder Way
You've got 27 coin, each of them is 10 g, except for 1. The 1 different coin is 9 g or 11 g (heavier, or lighter by 1 g). You should use balance scale that compares what's in the two pans. You can get the answer by just comparing groups of coins.
What is the minimum number weighings that can always guarantee to determine the different coin.
Separate the coins into 3 stacks of 9 (A, B, C). Weigh stack A against B and then A against C. Take the stack with the different weight (note lighter or heavier) and break it into 3 stacks of 3 (D, E, F). Weigh stack D against E. If D and E are equal, then F is the odd stack. If D and E are not equal, the lighter or heavier (based on the A, B, C comparison) is the odd stack. You now have three coins (G, H, I). Weigh G and H. If G equals H, then I is the odd and is lighter or heavier (based on the A, B, C comparison). If G and H are not equal, then the lighter or heavier (based on the A, B, C comparison) is the odd coin.
If you like this type of brain teasers, then surely check out other weighing puzzles. There are many easier ones as well.
You are one of 20 prisoners on death row with the execution date set for tomorrow. Your king is a ruthless man who likes to toy with his people's miseries. He comes to your cell today and tells you:
“I’m gonna give you prisoners a chance to go free tomorrow. You will all stand in a row (queue) before the executioner and we will put a hat on your head, either a red or a black one. Of course you will not be able to see the color of your own hat; you will only be able to see the prisoners in front of you with their hats on; you will not be allowed to look back or communicate together in any way (talking, touching.....).
The prisoner in the back will be able to see the 19 prisoners in front of him. The one in front of him will be able to see 18…
Starting with the last person in the row, the one who can see everybody in front of him, he will be asked a simple question: WHAT IS THE COLOR OF YOUR HAT?
He will be only allowed to answer “BLACK” or “RED”. If he says anything else you will ALL be executed immediately.
If he guesses the right color of the hat on his head he is set free, otherwise he is put to death. And we move on to the one in front of him and ask him the same question and so on…
Well, good luck tomorrow, HA HA HA HA HA HA!”
Now since you all can communicate freely during the night, can you find a way to guarantee the freedom of some prisoners tomorrow? How many?
First guy is a coin toss - let's wish him good luck.
His job is to establish the parity of black hats visible to him.
He says "Black" if he sees an odd number of black hats; "Red" otherwise.
By paying attention to what has been said, each prisoner will know his hat's color.
Example:
Second to speak hears "Black" and sees an even number of black hats.
He knows his hat is black [odd changed to even - must be his is black] and says "black".
Third guy has heard "black" and "black" and sees an even number of black hats.
He knows his hat is red [even stayed even - his hat can't be black] and says "red".
And so on, to the front of the line.
General algorithm:
The first time you hear "black", say to yourself "odd".
Each time your hear "black" after that, change the parity: "even", "odd", ... etc.
When it's your turn, if the black hats you see match the running parity, you're Red; Black otherwise.
Call out your color.
If you like this type of brain teasers, then surely check out other logic puzzles. Especially "Masters of Logic Puzzles" series at the bottom of that page might be of interest for you.
3. The Liar, The Truth Teller and The Random Answerer
There is a truth teller (always tells the truth), a liar (always lies), and one that sometimes answers truthfully and sometimes lies. Each man knows who is who. You may ask three yes or no question to determine who is who. Each time you ask a question, it must only be directed to one of the men (of your choice). You may ask the same question more than once, but of course it will count towards your total.
What are your questions and to whom will you ask them?
There are 6 possible states for the order of the men: TRL, TLR, LTR, LRT, RTL, RLT
There are 8 possible combinations of anwers for questions: TTT, TTL, TLT, TLL, LTT, LTL, LLT, LLL.
Theoretically it's possbile if you could figure out a way to get any of the 8 combinations of answers assigned to the states, but with the unreliability of Random's answers, I thought it was impossible. There is always a possiblity in any solution where Random will exactly mirror T or L for answers. He could always lie or always tell the truth and you can never tell when he is lying or telling the truth. This being given, I thought you can NEVER separate 6 distinct answers to apply to the 6 states, and therefore can never be sure who is who.
After a minute though, I saw through my own error in logic. I was always dealing with questions where T and L would give the same answer regardless of the order of the men. I saw that if you can get T and L to give a Yes/NO answer, then you can figure out where R's worthless answers are. The only way I saw to do this is to ask about the order of the men themselves.
So:
Ask #1 if L is standing on R's right arm (our left if they are facing us).
The answer gives you a split in the order they are standing:
If YES, then it has to be T telling the truth, L telling a lie, or one of R's worthless answers, so: TLR, LTR, or RTL, RLT.
If NO, then it has to be T telling the truth, L telling a lie, or R and his worthless answers, so: TRL, LRT, or RTL, RLT.
Now we know, based on the answer to #1 where to avoid R's worthless answers. We now ask T or L "Is T in the lineup?" If answer 1 was Yes, we ask person 2, if it was no we ask person 3.
The answer now will give us some more info. If it's Yes, it's T answering the truth, if it's no, it's L answering a lie. So based on who we asked, we now know:
Yes, Yes: Has to be LTR, or RTL
Yes, No: TLR, RLT
No, Yes: LRT, RLT
No, No: TRL, RTL
Now any question separating the two possiblities works - just make sure you are avoiding R's worthless answers.
For example:
Yes, Yes - ask #2 if #1 is L. (We know #2 is T and will tell the truth) - Yes = LTR, No = RTL
Yes, No, - ask #2 if #1 is T. ( We know #2 is L and will tell a lie) - Yes = RLT, No = TLR
No, Yes - ask # 3 if #1 is L. (We know #3 is T and will tell the truth) - Yes = LRT, No = RLT
No, No, - ask #3 if #1 is T. (We know #3 is L and will tell a lie) - Yes = RTL, No = TRL
So we have the order and know who is who.
If you like this type of brain teasers, then surely check out other logic problems. There are many easier ones as well.
Three boxes are all labeled incorrectly, and you must get the labels right.
The labels on the boxes read as follows:
[box 1] nails
[box 2] screws
[box 3] nails and screws
To gain the information you need to move the labels to the correct boxes, you may remove a single item from one of the boxes. You may not look into the boxes, nor pick them up and shake them, etc.
Can this be done? If so, how? If not, why not?
Remove an item from box 3.
The item tells you what label to put on box 3.
Move the nails and screws label to the box labeled with the other item, and its label to the remaining box.
Example: you remove a nail from box 3.
Move the label nails from box 1 to box 3.
You can't move the nails and screws label to box 1: that would be a swap, and all three labels must be corrected.
Move it instead to box 2, and the screws label to box 1.
[box 1] screws
[box 2] nails and screws
[box 3] nails
Teanchi and Beanchi are a married couple (dont ask me who he is and who she is)! They have two kids, one of them is a girl. Assume safely that the probability of each gender is 1/2.
What is the probability that the other kid is also a girl?
Hint: It is not 1/2 as you would first think.
Of course, it's not 1/2 else would make it a lousy puzzle.
Ans: 1/3
This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.
The following are possible combinations of two children that form a sample space in any earthly family:
Girl - Girl
Girl - Boy
Boy - Girl
Boy - Boy
Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.
This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3
You are on a game show and there are three doors. The presenter tells you that behind one of doors there is a car and behind the other two are goats. If you pick the car you win it. After you have picked a door the presenter opens a different door with a goat behind it, he then gives you the chance to change what door you open. What should you do?
Hint: It is not 1/2 as you would first think.
Jennifer should switch. Contrary to what may seem intuitive, switching actually doubles her chances of winning the car.
This problem is just a re-wording of what is known as the Monty Hall Problem. The key to understanding it is that the host knows the locations of the car and goats. His knowledge changes his actions and thus affects the odds.
Here is a breakdown of all the possible scenarios that Jennifer faces and why Jennifer should switch:
Door #2 has goat B (probability 1:3) - MB shows goat A behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances (1:3 x 1:1 x 1:1 = 1:3)
Door #2 has the car (probability 1:3) - MB shows goat A behind Door #1 (1:2) - goat B is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)
Door #2 has the car (probability 1:3) - MB shows goat B behind Door #1 (1:2) - goat A is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)
There are (1:3 + 1:3 = 2:3) chances that switching will get Jennifer the car, and only (1:6 + 1:6 = 1:3) chances she would get the car by not switching. She should switch.
A more general presentation of the reasoning is this:
At the start of the game, there is a 2:3 chance that Jennifer will pick a door with a goat behind it. If she does, the host will reveal the other other goat, and switching doors will get Jennifer the car.
There is a 1:3 chance she will pick the car. The host will then reveal a goat. Switching would win Jennifer a good supply of Ch?vre (and the disdain of her neighbours).
So, 2 out of 3 times switching gets the car. Simple - unintuitive, but simple.
Why does the host's knowledge change the odds. Because he does not randomly select a door to open - he always opens a door with a goat. By doing this he reduced the possible scenarios for Jennifer to the four listed above. If he randomly picked, then Jenny's chances, if the show progressed as presented, would be 50/50. However, there would also be a 1:3 chance that MB would open the wrong door and reveal the car's location (followed by a 1:1 chance that MB would be sacked and re-runs of McGyver would fill out the remainder of the season!)
Question 1: What is the next two rows of numbers?
Question 2: How was this reached?
The secret is to say whay you see on each line and what you see becomes the next line. For full answer see below.
Line 1 is "Two ones" (2 1)
Line 2 then becomes "One two, and one one" (1 2 1 1)
Line 3 therefore is "One one, one two and two ones" (1 1 1 2 2 1)
Line 4 is "Three ones, two twos and one one" (3 1 2 2 1 1)
Line 5 is "One three, one one, two twos and two ones" (1 3 1 1 2 2 1 1)
A man runs a mile south, a mile west, and a mile north and ends up back where he started!
But the real riddle is that there are actually an infinite number of answers for where the man could have started from. Explain.
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.
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.
A 6-inch hole is drilled through a sphere. What is the volume of the remaining portion of the sphere?
Clarifications:
[1] the hole is a circular cylinder of empty space whose axis passes through the center of the sphere - just as a drill would make if you aimed the center of the drill at the center of the sphere and made sure you drilled all the way through.
[2] the length of the hole [6 inches] is the height of the cylinder that forms the inside surface once the hole is drilled. picture the inside surface as viewed from inside the hole and measure the length of that surface in the direction of the axis of the drill.
in this sense, you could for example drill a 6-inch hole through the earth. the diameter of the hole would be huge, and you'd just have a tiny remnant of the earth left. but if you could set it on a table [a big table] it would be 6 inches high.
You of course could not drill a 6-inch hole through a sphere whose diameter was less than 6 inches. This fact leads to the logical answer.
The hard way involves calculus. The easy way uses logic.
Here's the mathematical solution:
Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h.
Since r*r = R*R - L*L and h = R-L, we can eliminate r and h and do everything in terms of R and L.
Swizzling cpotting's cap formula, V[cap] = pi/3 (2R*R*R - 3R*R*L + L*L*L)
Cylinders are ho-hum, V[cyl] = 2pi*L*r*r = (2pi/3) (3R*R*L - 3L*L*L)
V[removed by drilling] = V[cylinder] + 2V[cap]
doing the math,
V[removed] = (4pi/3)R*R*R - (4pi/3)L*L*L
Pretty amazing: the volume removed by a hole of length 2L is the difference of the volumes of two spheres: one of radius R, the other of radius L.
So the remaining volume is simply the volume of a shpere with radius L. [hint-hint at the logical solution]
V[remainder] = (4pi/3)L*L*L = 36pi.
Here's the logical solution:
My friend wouldn't have posed a math problem [boring] and she wouldn't have left out critical information. Therefore the answer couldn't depend on the radius of the sphere. I chose a sphere size [radius=3] that would make 0 volume removed [a hole of length 6 and diameter 0]. With nothing removed, the remaining volume is the original volume: (4pi/3) 3*3*3.
So it's 36pi.
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.
Find the smallest number with such property.
2519
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.
There is a room with no windows, doors, or any sort of opening, the walls are solid steel 10 feet thick, and you are trapped inside, left only with a saw and a table.
How do you escape? (more variations in the forum)
Fantasy logic problem where words have alternate meanings (usually related to homonyms)
You saw the table in half; but the two halves together; two halves make a whole (hole) and you crawl out through the hole.
As I was going to Saint Ives,
I crossed the path of seven wives.
Every wife had seven sacks,
Every sack had seven cats,
Every cat had seven kittens,
Kittens, cats, sacks, wives,
How many were going to Saint Ives?
It is 1
You are PASSING those people. You are the ONLY person going from X to St Ives, they are going from St Ives to X
1 26 L of the A - 26 Letters of the Alphabet
2 7 D of the W - 7 days of the Week
3 7 W of the W - 7 wonders of the world
4 12 S of the Z - 12 signs of the zodiac
5 66 B of the B - 66 books of the bible
6 52 C in a P (W J) - 52 cars in a pack (without jokers)
7 13 S in the U S F - 13 stripes in the united states flag
8 18 H on a G C - 18 holes on a golf course
9 39 B of the O T - 39 books of the old testament
10 5 T on a F - 5 toes on a foot
11 90 D in a R A - 90 degrees in a right angel
12 3 B M (S H T R) - 3 blind mice (see how they run)
13 32 is the T in D F at which W F - 32 is the temperature in degrees ferenheight at which water freezes
14 15 P in a R T - 15 players in a rugby team
15 3 W on a T - 3 wheels on a tricycle
16 100 C in a D - 100 cents in a dollar
17 11 P in a F (S) T - 11 players in a football (soccer) team
18 12 M in a Y - 12 meters in a yard19 13 is U F S - 13 is unlucky for some
20 8 T on an O - 8 tenticles on an octopus
21 29 D in F in a L Y - 29 days in Feburary in a leap year
22 27 B in the N T - 27 books in the new testiment
23 365 D in a Y - 365 days in a year
24 13 L in a B D - 13 loaves in a baker's dozen
25 52 W in a Y - 52 weeks in a year
26 9 L of a C - 9 lives of a cat
27 60 M in an H - 60 minutes in an hour
28 23 P of C in the H B - 23 pairs of chromosomes in the human body
29 64 S on a C B - 64 squares on checkers board
30 9 P in S A - 9 provinces in south africa
31 6 B to an O in C - 6 bowls to an over in cricket
32 1000 Y in a M - 1000 years in a millenium
33 15 M on a D M C - 15 men on a dead man's chest
What is your favorite puzzle? The best you've ever head? Let us know and post it as new topic into our New Puzzles Forum.
Would you like some easier brain teasers? Check out the following cool math games or logic puzzles
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