Math Credit Challenge

[MagicCarpetRide]

Re: Math Credit Challenge

you may as well have said what's 5+3, except the answer to yours can be typed more quickly

Actually they could be typed just as fast....

[Go_Oilers_Go]

Re: Math Credit Challenge

If two people are having sex and a third one joins in, how many people are there?

Think carefully...

^_^

[pokelda]

Re: Math Credit Challenge

thread sucks

[silvercomet1525]

Re: Math Credit Challenge

How about the Riemann Hypothesis?

...

Okay, is all seriousness... what is the next number in this sequence:

6, 35, 143, 323, 667, 1147, 1763, 2491, 3953, 5183...

Ten numbers should be enough to get rid of ambiguity of the sequence.

[Reach]

Re: Math Credit Challenge

How about the Riemann Hypothesis?

...

Okay, is all seriousness... what is the next number in this sequence:

6, 35, 143, 323, 667, 1147, 1763, 2491, 3953, 5183...

Ten numbers should be enough to get rid of ambiguity of the sequence.

D:

It should proceed, after 2491...3599, 4757, 5767...etc.

You missed a value >_>...!

[dAnceguy117]

Re: Math Credit Challenge

Actually they could be typed just as fast....

I type letters faster than I type numbers haha

[ledwix]

Re: Math Credit Challenge

A runner sprints around a circular track of radius 100.0 meters at a constant speed of 7.00 meters per second. The runner's friend is standing at a distance of 200.0 meters from the center of the track. How fast is the distance between the friends changing when the distance between them is 200.0 meters?

[silvercomet1525]

Re: Math Credit Challenge

D:

It should proceed, after 2491...3599, 4757, 5767...etc.

You missed a value >_>...!

Okay... maybe we're not thinking of the same thing? I kind of made this up, but it does have mathematical logic behind it.

[MrRubix]

Re: Math Credit Challenge

Reach: 56980, found manually. Working on the formula to make sure it accounts for every case.
EDIT: Let X denote the sum and let sub-n denote the first n elements to sum: Xn = [sum from i=1 to i=n(n+1)/2] i^2

So if you wish to find X(3), that means you sum from i=1 to 3*(3+1)/2 or 6, so you sum the first 6 squares: 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2, or 91, which is indeed equal to the first 3 terms summed. Verify with X(10): sum from i=1 to i=10(10+1)/2 or 55. The sum of the first 55 squares is equal to 56980.

Credits!

Tass: LOL if I solved p=np I'd want a lot more than FFR credits.

[MagicCarpetRide]

Re: Math Credit Challenge

A runner sprints around a circular track of radius 100.0 meters at a constant speed of 7.00 meters per second. The runner's friend is standing at a distance of 200.0 meters from the center of the track. How fast is the distance between the friends changing when the distance between them is 200.0 meters?

There is no such thing as a circular track.
I win.

[Reach]

Re: Math Credit Challenge

Okay... maybe we're not thinking of the same thing? I kind of made this up, but it does have mathematical logic behind it.

No, you made a mistake. To be more specific, IIRC you skipped the prime 61 after 59 and went straight to 67.


And well done Rubix. Yea, the problem is designed such that it can be solved without any expertise in mathematics actually.


This one is tricky.

Assume that gravity on or near the airless surface of a certain planet called Reach is such that any free-falling object will reach a velocity of 10 meters per second after one second, 20 meters per second after two seconds, 30 meters per second after three seconds, and so forth. Suppose a frictionless pulley (i.e., a wheel with a grooved rim that, once in motion, does not slow down due to friction), having negligible mass, is attached to a sturdy beam at the top of a vertical, airless mine shaft, and an inelastic rope 1000 meters in length but of negligible weight is threaded through the pulley with identical buckets of negligible weight dangling at each end. In one bucket are fifty stones weighing 1 kilogram apiece and an alien named Rubix weighing 50 kilograms, while in the other bucket is a single stone weighing 50 kilograms and an alien named Keeley weighing 50 kilograms. Suppose Rubix commences dropping stones out of its bucket at a rate of one per second. The instant the first stone is dropped Keeley commences climbing the rope directly above its own bucket at a rate of one meter of rope per second. If Rubix starts at rest at a position 20 meters below Keeley, how many stones will Rubix have dropped the instant the two buckets are precisely side by side?

edit: I gave you 1000 credits Rubix.

[MrRubix]

Re: Math Credit Challenge

I see the pattern too in silvercomet's sequence and I agree -- it starts to not work.

I know how to do the pulley problem too I think -- I love these kind of physics problems. Will solve it when I get back (picking up some drinks)

[silvercomet1525]

Re: Math Credit Challenge

Okay, I guess it was a bit obvious to you huh? Yeah, you got the pattern, I just made a mistake.

[albaneenesk8r]

Re: Math Credit Challenge

mr rubix + FFR = ______

[Reach]

Re: Math Credit Challenge

I see the pattern too in silvercomet's sequence and I agree -- it starts to not work.

I know how to do the pulley problem too I think -- I love these kind of physics problems. Will solve it when I get back (picking up some drinks)

I will be impressed if you can! It's quite the mental gym. Though there are harder problems surely (I'm sure I could make that same problem drastically harder, but then I wouldn't know how to solve it myself XD), I've never met anyone that could actually solve it...at least without a second try.