You think you're a maths nerd?

[arjue]

I just saw these posted at another forum, and thought they were interesting. give them a try, if you dare.

a professor met one of his brighter students on campus one morning with the following riddle:

"i met 3 persons last night. the product of their ages was 2450. the sum of their ages was twice your age. how old were they?"

the student, after some thought, replied "i dont belive you have given me enough information"

the professor responded "that is correct, i have not given you enough information. let me add that i am older than any one of the three. now you have enough information"

a homeless guy in the bushes overhearing this conversation jumped up and remarked "i can tell you their ages, furthermore, i know both of your ages as well"

how old are the 3 people, the student and the professor?

You are given 12 marbles which appear to be exactly the same. ONE of these twelve is slightly different from the others. It is either lighter or heavier, but you don't know which.

You have a two sided balance scale (nothing funky just a normal two armed balance scale) with which you are allowed to make THREE measurements.

What three measurements (groupings of marbles) are required to determine which marble is different and whether it is heavier or lighter than the others?

note: I couldn't figure out the solutions myself, but they were posted by other people.

[MVS]

The marble thing, I know you do 1 2 3 4 V 5 6 7 8 for the first one.

if that's even, then you know it's 9-12 and use the 2 turns to figure that out.

If it's not even, you know it's 1-8 and that's about where it all falls to shit for me.

[Timber]

This sounds like the questions we got in Math League back in HS. It's been 4 years since I did any math more complicated than balancing my checkbook and I'm a goddamned pothead so I didn't think I'd get 'em. I got about as far as Matt on the marble one (although backwards) and I can't even remember what "product" refers to (multiplication?)

[Melf]

I'm an English teacher and thankfully, the only math I do is with a calculator!

[Rich]

For the second one, do 6 marbles vs 6 marbles, then 3 marbles vs 3 marbles, then 2 marbles vs 1 marble and hope it's the one marble that's different. If not, take a 50% chance at guessing.

[Hero]

the marble one sounds a lot like a problem I had to do in HS...

I wish I could remember how it went

Let's try this: Take marbles 1-4 on one side, 5-8 on the other. If the weights are equal, you know 9-12 has the heavier marble. Or, if one side is heavier, you know it that way. Take the set that's the heaviest (lets say 1-4 is) and split it in half (turn 2). Take the side that's heaviest from that, split it in half (one marble on each side) and you know the answer (and in three turns).

[Captain Vegetable]

Originally posted by Rich
For the second one, do 6 marbles vs 6 marbles, then 3 marbles vs 3 marbles, then 2 marbles vs 1 marble and hope it's the one marble that's different. If not, take a 50% chance at guessing.

Actually, it's 6 v. 6, 3 v. 3 (the more heavy group of 6), and then 1 v. 1 (from the more heavy group of 3). If the last two you weigh are balanced, you know the one you didn't weigh is the screwy one. If the two don't balance, you know which is the screwy one.

Oh, wait. We don't know if it's heavier or lighter...hmmmmm.

[radiant]

we seem to be forgetting that the marble is either heavier or lighter. rich, hero, vegetable, your methods wouldn't work. :/

[Captain Vegetable]

Originally posted by radiant
we seem to be forgetting that the marble is either heavier or lighter. rich, hero, vegetable, your methods wouldn't work. :/

I know. I edited.

[radiant]

heh, just beat me.