Vivek
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4?
Suppose there are n photos. There are 8 friends. Each photo has 5 friends. So, Total count of people in photos is 5n. Also Let x friends appear in 2 photos and (8x) friends appear in 3 photos. So, Total count is 2x + 3(8x) = 24x. So, 24x=5n. Note that x<=8. So, only possibility for 24x to be the multiple of 5, is when x=4. That gives 5n=20 and so, n=4.

The distance from the base of the lighthouse to the light beam is not going to change as it revolves and it is sqrt(x^2+y^2). So The speed is 2*pi*sqrt(x^2+y^2) per minute.

When n is even, It has no real solution. When n is odd, It has exactly one solution. We can prove that by induction. Assume that this is the case when n=2k and n=2k+1. Consider the case when n=2k+2. It is easy to see this curve lies above the x axis, based on differential tests and the fact , f'(n,x)= f(n1,x). And When n=2k+3. It must have at least one solution. If it has more than one, then by roll’s theorem f(2k+2,x)=0 will have at least one solution, leading to contradiction. And That proves the induction.

Let xi be the number of times the number “i” appears in the sequence.
So, We need to find the number of solutions for the equation
x1+x2+x3+…+xN = M where xi>=0
And That is ((M+N1) choose (N1)).

Let us name any two corners based on their position relative to each other.
Consider a corner, say A.
It has 3 adjacent corners, 3 corners that are diagonally opposite but lies on the same side as A, 1 corner which is exactly at the opposite corner.
Let the expectation to reach the same corner be S, to reach the adjacent corner be A, to reach diagonally opposite corner on the same side be D, exactly opposite corner be P.
So,
S=1/3*(A+1)+1/3*(A+1)+1/3*(A+1)
So, S=A+1
A=1/3*(1)+1/3*(D+1)+1/3*(D+1)
So, A=2D/3+1
D=1/3*(A+1)+1/3*(A+1)+1/3*(P+1)
So, D=2A/3+P/3+1
P=1/3*(D+1)+1/3*(D+1)+1/3*(D+1)
So, P=D+1
Solving these, gives A=7,S=8,D=9,P=10.

For the first question, Is it 3/5 ? And May you explain the second question?

70 degrees. Using sine and cosine law of triangles, and standard trigonometric identities.


Using Indicator variables. And the ending point situations have probability of 1/2 to be local minimum and others have 1/6. So, 2*(1/2)+98*(1/6) =52/3 =17.3333 ?

990/2^10 ? I tried using indicator variables. But not sure of the answer. Is there any other way?

Interesting (When worlds collide?) it is. I too think about it. I am not sure whether I miss something. Wondering about this : Conduct the fair lottery. If only all the individuals in the group win, That group will be selected. Otherwise the entire group will be rejected. If none of the group wins, Conduct the lottery again ?

Usually the standard explanation goes like this : Consider for example , X1+X2+X3=5. The solution can be considered equivalent to 5 *s separated by 2 s . For example (1,2,2) corresponds to ***** . In the general case, M *s and N1 s and that gives ((M+N1) choose (N1))

Before your “pretty solution clue”, I got that through lots of equations , changing 1/2 to sin 30 and the like. haha..This is cool! ðŸ™‚

Another interesting question with the interesting answer as well! What is the expected number of throws till somebody wins?

It really is (n/(n+1))^n for n sided dice, approximating to 1/e as n increases. Confirmed it by simulating in R program. This is the nice way to see the e value with probabilistic simulation. I remember that the probability of derangement approaches 1/e. But This one is really super cool!