Quote:
Originally Posted by psyang
Ok, another puzzle.
You have a meter stick placed so that the 0cm marker is on the left, and the 100cm marker is on the right.
99 ants are placed on every cm marker from 1 to 99.
The ant at cm 1 is facing to the right. The ant at cm 99 is facing left.
For the 97 ants in between, a coin is flipped to determine if that ant is facing left or right: heads, the ant faces left, tails the ant faces right.
Ants only walk in a straight line at 1cm/s in the direction they are heading. If an ant bumps into another ant, both ants immediately turn around and start walking in the opposite direction.
If an ant walks to either end of the meter stick, it falls off.
1) Prove that, eventually, all ants will fall off?
2) What is the maximum time it will take for all ants to fall off?
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1. Horribly rough, but based on the way it works, because the ants are expected to keep moving till they fall off, Ant 1/99 will guarantee fall off after once bounce, 2/8 after 2 bounces etc.
It doesn't really matter which directions the other ants are facing because 1/99 are facing in and guarantees the next ant bounces and does a chain reaction before falling.
2. I want to say you calculate by determining the expected movements of ant 49 which is supposed to fall off last. I assume it bounces 49 or 48 times with the pairings, so add up 48cm + 47 till 1 for distance traveled in cm and then adjust for seconds/minutes/hours?
I'm sure someone else would get the actual accurate values or something.
Geez, these puzzles are like much more complex than elementary school math. Anything over elementary school math and it ceases to be fun. (ignoring higher level math tools that are just expansions of basic BEDMAS).
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The only math puzzle I distinctively remember was from like grade 3 in a "problem solving" math question. It popped up again in grade 6, so when I was the only person to answer it correctly.
I remember it distinctly because I think we were given like 20 minutes to do it and if you finished early, you went for recess. I handed in my paper 3-4 minutes after we started (teacher actually scolded me in front of the class because she thought I gave up and wanted extra recess) and I was the only one to get it correctly. To "fix" her error in scolding me, she gave me the "honor" of presenting my response to the whole class.
You have a machine that requires exactly 8 litres of liquid or something catastrophic will happen (trap, spaceship explodes, whatever). You have two oddly shaped containers that hold 5 and 7 litres exactly. These are the only two containers that are available that can hold the liquid and you have access to unlimited liquid.
How do you get the exact 8 litres using these two containers?