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# 334

How to drive a mathematician crazy - November 9, 2010
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Source for text: Missing Dollar Riddle

ahahah old jokes... learned this around 4 or 5 years ago...
obviously the 27 paid account for the 2 that the bellhop took, the missing 3 dollars are the ones that the guests were given back :P

well done :)

I love how she keeps on telling the riddle, completely ignoring his responses.

yep that's what happens with gals... they're so determined to find you dumb founded even at the cost of their dumbness!!

LOL nice

The clerk kept it.

That was what I finally had to say when I was 8 and was told this. I was able to explain it, but I just sounded so precious to the adults who knew better than there was no answer.

I always loved when people come to me with that "problem". But now that I read the the "follow-up" section in Wikipedia's article I cannot wait for the chance to give that answer instead.

I've had exactly the same experience with explaining the Monty Hall problem.

yeah well that's all well and good, but... WHAT HAPPENS TO THE MISSING DOLLAR!? :)

I took it! Satisfied?

I do remember an old problem in one of Marilyn vos Savant's columns, when she claimed there was no solution:

You have 50 coins and ten envelopes. How do you put all the coins into the envelopes so that each envelope has a different number of coins in it?

I looked at it, and after less then a minute came up with two different classes of solutions.

If every envelope must contain at least one coin then it is impossible. This condition was probably assumed by the author but not phrased in the question - since the problem is that of "integer partitions" where we partition a positive integer into a sum of positive (i.e. nonzero) integers.

yes, if every envelop needs coin, then it is impossible.

easy :)
Place 41 coins in one envelope.
Fold this envelope, place it and a coin inside the next envelope.
Rinse and repeat.

muahahaha... outside of the box thinking is fun.

Excellent solution... You deserve a cookie!!

I agree--the two classes were nested envelopes and empty envelopes, neither of which was illegal according to the original statement of the problem. Adding more coins might work--but it did specify 50 coins.

outside the box thinking or inside the envelop thinking :P

It should also state that I am not allowed to put additional coins in there. But yes, at any rate, it's trivial (0,1,..,7,8,14).

I also like the solution with the folded envelopes.

and of course stipulate that you cannot cut coins up either :)

One empty envelope is okay. Two isn't, because the problem specifies a different number of coins in each envelope. But then, you don't have a problem putting 50 coins in 9 envelopes in the standard fashion, so one is enough.

perhaps she was holding a diode?

Not a mathematician, but I've always found this "puzzle" annoying. I mean, it's so obvious what's wrong and there is absolutly no twist in it. Very boring...

How to drive someone (especially a cute girl) who thinks she's found a cool math riddle for you and is enthusiastically trying to tell it to you crazy: Blurt out the answer before she gets the first sentence out of her mouth and tell her dismissively that you've heard it a million times before.

C'mon guys, at least humor her. This is why people hate nerds.

Since when do people hate nerds?
Well, other than high-school, where everyone hates everyone.

at our high school it is standard to be a nerd. some more so than others (like deriving the true formula for the period of a pendulum as part of a lab write-up for a freshman physics class? and then showing how it approximates the given formula?)

NNNNNEEEEEERD :)

Yes--see tomorrow's version...

I'm gonna bookmark this page. I think of it as sort of tired-old-joke inoculation. 5 yrs from now if someone tries to spring this joke on me, I'm covered.

--
Furry cows moo and decompress.

I love the part where the blonde should just shut the hell up

"two dogs are running toward each--"

Me: 25 kilometers

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