M57 wrote:

I must be missing something..

Right

Do it faster than 19 min.

Is there out of bounds thinking involved?  i.e. toss the flashlight over?  Fill in the bridge gaps?

Ozyman wrote:

Is there out of bounds thinking involved?  i.e. toss the flashlight over?  Fill in the bridge gaps?

Nope - not needed.

Two more solutions... probably neither allowed.

How long is the bridge & how far does the flashlight throw a beam?   Can someone stand on the edge and still light the way for others to pass?

>They realize that at most two of them can be on the bridge at any given time, else it will collapse.

Torque on the bridge is proportional to the person's weight * their distance from the edge.   Downward force on the bridge is dependent on the ratio of their distance from each edge to the total bridge distance.   Their has got to be a sneaky answer in their somewhere.

Comon Ozy, think about it.  If the really slow guy walks at least half way with the second slowest guy, their cutting down the next slowest guys time addition by half. 

Ozyman wrote:

Two more solutions... probably neither allowed.

How long is the bridge & how far does the flashlight throw a beam?   Can someone stand on the edge and still light the way for others to pass?

>They realize that at most two of them can be on the bridge at any given time, else it will collapse.

Torque on the bridge is proportional to the person's weight * their distance from the edge.   Downward force on the bridge is dependent on the ratio of their distance from each edge to the total bridge distance.   Their has got to be a sneaky answer in their somewhere.

Nope - it's a simple solution.  Quite simple once you figure it out.  So you just have to tell me which two go first, who comes back, etc, etc...  and you can get them all across in less than 19 min.

Just don't try every single combination (brute force) - that would be cheating.  It's a duel optimization problem - so you need to minimize on both the forward trips as well as the return trips.  At least, that's how it makes sense in my mind when I'm thinking about it.  I can also think of it as a heuristic, rule-based, problem.

Oh, and as far as the bridge collapsing, any two people can be on it.  The instant a third person puts a foot on it, it drops.  And the bridge is far too long for the light to reach the other side.

Is it possible to use a rope ?

Ok, I am not gonna use a rope to get the flashlight back without crossing.

I think I've got it now, but it was hard I reckon. That was a good one Slander.

--> AB : 2 min

<-- A   : 1 min

--> CD : 10 min

<-- B   : 2 min

--> AB : 2 min

So it makes 17 minutes instead of 19.

Toto wrote:

Ok, I am not gonna use a rope to get the flashlight back without crossing.

I think I've got it now, but it was hard I reckon. That was a good one Slander.

--> AB : 2 min

<-- A   : 1 min

--> CD : 10 min

<-- B   : 2 min

--> AB : 2 min

So it makes 17 minutes instead of 19.

Yep - that's it.  The trick is to get the 5 and 10 min to go together without having to make a return trip (optimize the forward trip), while still trying to use the fastest available person to return the flashlight (optimize on the return trip).  Once you realize that that implies that the 5 and 10 min need to go together on the 2nd crossing, everything else falls into place.

Decapitate me, and all becomes equal.  Truncate me, and I become second.  Cut me front and back, and I become 2 less than I started.

What am I?

Switching light bulbs:

You have 100 light bulbs, each controlled by it's own switch.
They are numbered 1 to 100
All switches start in the off position.

On Pass #1, go to every switch that is a multiple of 1 and flip the switch (hint: this turns all bulbs on).
On Pass #2, go to every switch that is a multiple of 2 and flip the switch (that means, flipping switch for bulb 2, 4, 6....etc)
On Pass #3, go to every switch that is a multiple of 3 and flip the switch (3, 6, 9....etc)

Repeat this incremental pattern 100 passes, each time flipping switches (from off to on, and on to off) for the switches that are the multiple of the pass number.

(Preferably without using the brute force method) When finished with all 100 passes, how many bulbs will be left on, and which bulbs will they be?

The primes will turn off.

All Square numbers will will be left on. E.g. 7 off 49 on.

I'm not sure, but I'm thinking all of the composite numbers will have factor pairs that will cancel each other out, for instance take 24: 

1 on, then (2,12) (3,8) (4,6), and 24 should turn it off.

10 squares (1 is not a prime but is a square) = 10 left on.

That's my best guess first thing in the morning ..with an espresso shot to help.

That sounds right. For it to be on, it needs an odd number of unique factors which only happens when one of the factor pairs duplicates a number, thus the 10 square numbers between 1 and 100.

Bravo.  Well played.

Amidon37 wrote:

From a joke book I had as a kid - not sure if the riddle is well known or not, but I remember it fondly -

 

What can go up a chimney down, but can't go down a chimney up?

Looks like nobody answered this one.

It's an umbrella.

Goes up the chimney down (closed umbrella), but won't go down the chimney up (open umbrella).

A deceptively simple riddle:  How far can a dog run into the woods?

1/2 way

Brothers or sisters have I none, but that mans father is my fathers son. Who is that man? 

my son

Rich people need it, poor people have it, if you eat it you will die. What is it?