# Math problems

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• 02-04-2008, 06:49 PM
darkus-s6
the gravitational force on a mass inside a spherical shell = 0
when a mass m is inside a simmetric spherical distribution of mass, that part of the mass outside its radius doesn't contribute to the net force on it
• 02-04-2008, 09:33 PM
Correct, though I have no idea what you're trying to say in that second sentence.
• 02-06-2008, 11:18 AM
darkus-s6
^ that's the shell theorem

Ok, so here is another one:

13 musicians are playing a song. 6 of them are singing, 4 are playing tambourine, and 8 are playing some other instruments. How many are playing tambourine and some other instrument, if there are two musicians that only sing?
• 02-07-2008, 12:43 AM
Volvogga
You mean 18 musicians are playing a song?
If so, then 16... if not... uhh... well lets see:

Assuming no one is extremely talented with their feet, then we would conclude that tambourine and 'other instruments' take up the 'instrument' slot for a person. Beyond that, we assume microphone stands and what not for various 'other instruments' allowing one to play and sing at the same time. With that in mind, we first deal with tambourine and other, being 4 and 8 respectively. The sum of this is 12. There are 12 musicians out of 13, so one person must exclusively sing.

The new lineup is only 2 singers that ONLY sing, so we know the 1 must be the singer from before, and the other is most likely a dual singer/player again that has given up the instrument... making the total players one less than the previous total.

• 02-07-2008, 01:22 AM
irecinius
There are 13 playing
minus 2 that only sing, leaving us with 11 musicians that are playing any instrument (tambourine + others).

This would also mean one of them is dual plyaing the tambourine and other instrument, and maybe even singing at the same time.
• 02-07-2008, 02:44 AM
darkus-s6
Yes there is only one musician playing tambourine, an other instrument and possibly singing too, you're right irecinius
• 02-07-2008, 10:07 AM
Volvogga
Oh, hahahaaha. Damn, good one.
I read that completely wrong. XD

Irecinius, all you buddy.

• 02-07-2008, 11:41 AM
irecinius
lol... its my turn then? i'm not very good with questions, well lets see~

ABCDE * 4 = EDCBA, find A, B, C, D and E where each is a unique integer from 0 to 9.
• 02-07-2008, 03:26 PM
Are you sure that that's possible?

I'm starting at the edge, so A could be 0, 1, or 2, since 4A cannot be > 9.

It can't be 0; then E would have to be be 5 (5*4=20), but that's impossible since even 9999*4 != 5xxxx

It can't be 1 since 4E != x1, no matter what E is.

It can't be 2 since E must then be 8 or 9, 8*4=32, 9*4=36, neither of which ends in 2.
• 02-07-2008, 03:40 PM
irecinius
Yes it is, I'll give that you started on the right track, but derailed midway...

 I can just give the answer if someone wants.
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