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Maths and Physics (Problems)


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AuthorMaths and Physics (Problems)
12?

I think 10
You think wrong.
Could u explain?
0s will come equal to number of multiples of 5

So upto 50, there are only 10 multiples of 5, can u please explain where did the 2 extra 0s came from?
25 and 50.
Resonance is clearly not working out for you.
5, 10, 15, 20, 25, 30, 35, 40, 45, 50 -> 10 multiples only

Don't u dare talk about my coaching institute!
And please don't provoke others by hurting their feelings! This was meant to be a peaceful forum thread, not a place to show-off half knowledge!
You still don't get it?
Fine. Sorry.

25 is 5 squared.
50 is a multiple of 25 and hence has two multiples of 5 as well.
So. The answer is 10 + 2
ok, thanks for clearing but dude you should not hurt others
Cheryl's birthday problem is so easy :/
10th graders deserve a better challenge.
3 particles A, B and C are situated at vertices of an equilateral triangle all move with same constant speed such that A always moves towards B, B towards C and C towards A. Initial separation between each of the particle is 'a'. O is the centroid of the triangle. What is the distance covered by particle A when it completes one revolution around O ?
Allright a fine one .....
If p and q are distinct prime numbers then the number of dsitinct imaginary numbers which are pth as well as qth roots of unity are
89 ... one revolution would mean what ? Buddy total time to reach centre would be 2d/v nd if by one revolution you mean to reach bacak at A then the question is wrong because here the system is all relatively moving nd no particle wll be at same position again
@90, has to be 0
one root will always be 1.
And the number you are looking for will be a vertice of a regular polygon with one vertice at (0,1)
two vertices cannot co-incide if p and q are distinct prime numbers
Sum of roots oc equations z^3 +2z^2+2z+1=0 and z^97+z^29+1=0 is equal to
@93
-2 + 0 = -2
Can any one try to find the sum of real roots? :S
Post 69, nr 1:
The rope will be 1/2/pi = 0.15915... meters over the equator. Funnily enough, the diameter of the sphere (in this case earth) does not matter.

Post 69, nr 2:
Lets say we start at 00:01. In 24 hours the long needle does 24 full turns. The short needle does 2 turns. They pass each other 24-2=22 times. However, if we start at 00:00 they are initially covering each other and does so again at the next 00:00. So the correct answer would be 23 times.


First one is good and that's the result which I think is very interesting, the size of sphere doesn't matter (so it would be the same result even for a bottle cap :D).

In the second one I asked for the exact times as hh:mm, but your answer is good for the question you all understood, 23 times because at hour 11 the minute would be 60 which would mean hour 12. (so neither 24 or 22 as other people said).
Except that on all watches the movement of the needles isn't continuous but discreet. So you will have coincidence at 11:59 *and* 12:00

00 h 00 min
01 h 05 min
02 h 11 min
03 h 16 min
04 h 22 min
05 h 27 min
06 h 33 min
07 h 38 min
08 h 44 min
09 h 49 min
10 h 55 min
11 h 59 min
that's right ;)
An upgrade to the problem:
Suppose on a clock instead of 12 hours there are N hours and instead of 60 minutes and 60 seconds there are M minutes and M seconds on the clock:

How many times will the hour, the minute and the second needles superpose?
I've got a nice (unoriginal but modified) one.

There is a place such that you can go 1km south, 1km west, and 1km north (sequentially), and end up at the same place. If a "place" is defined as 1 cm^2 (sq. centimeter just confirming), then how many such places are there on Earth? Radius of earth is 6,371 km, and make *marginal* assumptions wherever required.

Please don't post your answer to this question yet..just mail me. It's a nice question, and I think everyone should get to try this :)

I will mail the solution to whoever asks for it.
for _CrackpoT_:


89 ... one revolution would mean what ? Buddy total time to reach centre would be 2d/v nd if by one revolution you mean to reach bacak at A then the question is wrong because here the system is all relatively moving nd no particle wll be at same position again


the system of particles will meet eventually at time 2a/v that's right..but they'll have to perform infinite revolution around the centroid of that triangle..the question asks the distance covered when it has revolved an angle of 2 pie around the cenroid..(still moving)

the answer is known to everyone who gave jee mains this year .. -_-
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