Recent content by ddddd28

Hello, I am trying to design a mechanism that moves a piece towards the center of a disk by rotating a second disk as described in the picture below. There is a stationary disk which restricts the motion of the piece, so it can move vertically only. There is also a second disk that it is located...

Hello, Consider the next scenario: I wish to travel from point a to point b as fast as possible. Between the points there is a border. In the region from a to the border, I can move only with velocity v1, and after the border, I am allowed to move only with velocity v2. All the additional...

Hello, Recently, a solar power tower plant was founded next to where I work. Since it's the tallest object in the area, it's quite hard to miss it. But apart from that, every morning the reflected light is arranged in a hyperbolic- like way, as you can see in the picture. Does anyone have a...

Are we taking a test right now? I believe that, when not tested, one should not be satisfied with a sufficient solution, because the most important thing is not to provide a solution, but to learn something from the attempts and to improve oneself for the next problem

Your shortcut works well with factorials, what about the rest of the numbers?

You are right, but it still works. Of course, the amount of 2s only matters, because the rest of the factorization is reduced anyway.

To make my point fully clear, take 60. The factorization is 2^2*3*5. Therefore, there are 3*2*2= 12 factors : 1,2,3,4,5,6,10,12,15,20,30,60. Similarly, there are only 2*2= 4 odd factors: 1,3,5,15. So, the probability in this case is one third.

You misunderstood me. I meant that the number of factors of 25! is 23*11*7*4*3*2*2*2*2. Since we also need to find how many odd factors there are, we don't need to include 23 in the product of 25!/2^22.

The number of factors can be simply described as the product: (q1+1)(q2+1) ... (qn+1). q1, q2 ... qn are the number of times a prime appears in the factorization. To find the required probability is to calculate the product of 25! and the product of 25! divided by 2^q.

Just wanted to share a nice method for visualizing the number of factors. All the factors are, of course, solutions of the equation xy=d ,d is a natural number. Notice that the function cos(2πk) equals 1 only and only if k is an integer. Since x and y have to be integers, it is possible to...

You are right:sorry: It doesn't hold... I don't understand how I missed it.

I think that the problem's composer used n=25! not in vain. In other words, counting the factors of a factorial is very easy. Just look at it in a recursive perspecive: n factors 1! 1 2! 1 2 3! 1 2 3 6 4! 1 2 3 6 4 8 12 24 5! 1 2 3 6 4 8...

Hello, The problem I came up with deals with the structures that can be obtained by joining squares side to side or corner to corner. Specifically to this problem, structures, that are symmetrical to each other, are regarded the same. Ideally, I am looking for a formula that will tell how many...

Since we are talking about limits here, constants don't matter when the degree is smaller than the divided one's (k^n+1). Regarding the leading coefficient, it's obvious that it is 1.

First, I am thankful for introducing the discrete calculus, a subject that I was not awared of at all. It encouraged me to study it a bit, realizing it's such a powerful tool, and yet quite simple. Stephen Tashi, I think it is not necessary to express explicitly the terms of the "Taylor's"...