U2 has a concert that starts in 17 minutes and they must all cross a bridge to get there. All four men begin on the same side of the bridge. You must help them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them.
The flashlight must be walked back and forth. It cannot be thrown and other tricks like that are not needed to solve the problem. The solution is simply a matter of allocating resources in a certain order. Each band member walks at a different speed. A pair must walk together at the rate of the slower man’s pace:
Bono: 1 minute to cross
Edge: 2 minutes to cross
Adam: 5 minutes to cross
Larry: 10 minutes to cross
For example: if Bono and Larry walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Larry then returns with the flashlight, a total of 20 minutes have passed and you have failed the mission.
This is one of my favorite brain teasers and I want to give you the satisfaction of figuring it out on your own. If you’re having a hard time, here’s a hint: There is a valid answer that doesn’t require tricks like throwing the flashlight or shining it backwards or having some other means of moving the flashlight.
There’s an assumption people often make that keeps them from solving this. Two members cross the bridge each time, but neither one of the two who crossed necessarily need to return. Think about how that would be possible. If you’re still stuck, use objects to simulate their movements. Use whatever you have laying around – pens, paper, erasers – and move them back and forth. Good luck!
Most of you have two eyes. I only have one.
Most of you have eyeballs. I do not.
Your eyes aren’t dangerous, neither are mine.
But all together, I am extremely dangerous.
You can see things with your eyes.
I can’t see anything, even though the air is clear where my eye is.
A princess is as old as the prince will be when the princess is twice the age that the prince was when the princess’s age was half the sum of their present ages.
This one took a while to figure out and there are numerous valid ways of finding the answer.
Here is the solution I came up with
I created the following table from the riddle:
Current
Future
Past
Princess
x
2z
(x+y)/2
Prince
y
x
z
I then created three equations, since the difference in their age will always be the same.
d = the difference in ages
x – y = d
2z – x = d
x/2 + y/2 – z = d
I then created a matrix and solved it using row reduction.
x
y
z
1
-1
0
d
-1
0
2
d
.5
.5
-1
d
It reduced to:
x
y
z
1
0
0
4d
0
1
0
3d
0
0
1
5d/2
This means that you can pick any difference you want (an even one presumably because you want integer ages).
Princess age: 4d
Prince age: 3d
Ages that work
Princess
Prince
4
3
8
6
16
12
24
18
32
24
40
30
48
36
56
42
64
48
72
54
80
60
To see other solutions check out the comments from when I posted this on my blog.
Walking down the street one day, I met a woman strolling with her daughter. “What a lovely child,” I remarked. “In fact, I have a younger child as well,” she replied.
What is the probability that both of her children are girls?
1/2 probability. This has been know to cause raging debates and is known as one of the variations of the Boy or Girl paradox. This variation is more straightforward because knowing the position of the child leaves only two possibilities – the other child is a boy or a girl, each of which have a 1/2 probability.
Everly and I were playing on the merry-go-round at the local park. It was very large and we stood on opposite sides. As we spun the merry-go-round counter-clockwise, I threw a ball to Everly. Did the ball go to Everly, to the right or left of them?