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1) Let x be a random positive integer between 1 and 100 inclusive. Find the probability x leaves a remainder of 4 when divided by 7. (SOLVED) 2) Let a and b be random reals between 0 and 10 inclusive. Find the probability that a^2+b^2 =< 1 and y >= 1 - x. 3) 11 indistinguishable candies are randomly distributed to 4 kids. Find the probability that none of the kids have at most 2 candies. 4) Let a, b, and c be reals between 10 and -10 inclusive. Find the probability a^2+b^2+c^2 =< 1 and exactly one of the variables a, b and c are positive. 5) There are 6 dice numbered with the integers between 0 and 5 inclusive that are fair. Find the probability that when all of the dice are rolled simultaneously, the sum of the numbers shown on the dice is 15. (Its okay to express your answer as binomials, if you do happen to do so, use the form binomial(a, b) where binomial(a, b) defines the number of ways to choose a committee of b people from a people.) Thank you, SuperJupiter5 (Edited username)

How do find an analytic solution to this infinite series. (plug this into http://atomurl.net/math/ (latex)) \sum_{n=1}^\infty \csc^2(\omega\pi n)= \frac{A}{\pi} +B I have attempted the following to break the series down into 4 \sum_{n=1}^\infty \csc^2(\omega\pi n)= \sum_{n=1}^\infty csch^2(i\omega\pi n)= 4\sum_{n=1}^\infty \big(e^{\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big) }-e^{-\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big)}\big) ^{-2} \sum_{n=1}^\infty \big(e^{\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big) }-e^{-\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big)}\big) ^{-2}= \big(ie^{\pi\frac{\sqrt{3}}{2}}+ie^{-\pi\frac{\sqrt{3}}{2}}\big)^{-2}+ \big(-e^{\pi\sqrt{3}}+e^{-\pi\sqrt{3}}\big)^{-2} +\big(-ie^{3\pi\frac{\sqrt{3}}{2}}-ie^{-3\pi\frac{\sqrt{3}}{2}}\big)^{-2}+ \big(e^{2\pi\sqrt{3}}-e^{-2\pi\sqrt{3}}\big)^{-2} +... = \sum_{n=0}^\infty \big(ie^{(4n+1)\pi\frac{\sqrt{3}}{2}}+ie^{-(4n+1)\pi\frac{\sqrt{3}}{2}}\big)^{-2} +\sum_{n=0}^\infty \big(-e^{(2n+1)π√3}+e^{-(2n+1)π√3}\big)^{-2} +\sum_{n=0}^\infty \big(-ie^{(3+4n)\pi\frac{\sqrt{3}}{2}}+-ie^{-(4n+3)\pi\frac{\sqrt{3}}{2}}\big)^{-2}+\sum_{n=0}^\infty \big(e^{(2n)π√3}-e^{-(2n)π√3}\big)^{-2} \sum_{n=0}^\infty \big(-e^{(4n+1)\pi\sqrt{3}}-2-e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=0}^\infty \big(e^{2(2n+1)\pi\sqrt{3}}-2+e^{-2(2n+1)\pi\sqrt{3}}\big)^{-1} + \sum_{n=0}^\infty \big(e^{(3+4n)\pi\sqrt{3}}-2+e^{-(3+4n)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=1}^\infty \big(e^{4n\pi\sqrt{3}}-2+e^{-4n\pi\sqrt{3}}\big)^{-1} I have found the sums numerically and \sum_{n=0}^\infty \big(-e^{(4n+1)\pi\sqrt{3}}-2-e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=0}^\infty \big(e^{2(2n+1)\pi\sqrt{3}}-2+e^{-2(2n+1)\pi\sqrt{3}}\big)^{-1} + \sum_{n=0}^\infty \big(e^{(3+4n)\pi\sqrt{3}}-2+e^{-(3+4n)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=1}^\infty \big(e^{4n\pi\sqrt{3}}-2+e^{-4n\pi\sqrt{3}}\big)^{-1} \approx -0.00429 The only way I know how to evaluate the sum is using the polygamma function however it does not give values for A and B. Any help would be grand!

I know the answers, just want to share some interesting geometry facts, and questions are fun! 1) Which regular polygon contains (hidden within it) the golden ratio? 2) What SHAPE do you get when you add the volume of a hemisphere to that of a cone? (if all three shapes have same height and radius).

Sorry that I could not be active recently. I had some important family matters to deal with, because of the virus. Anyway, here is the problem: The 5 x 5 array of dots represents trees in an orchard. If you were standing at the central spot marked C, you would not be able to see 8 of the 24 trees (shown as X). If you were standing at the center of a 9 x 9 array of trees, how many of the 80 trees would be hidden?

ICE CREAM CONE VS SODA CAN You have an ice cream cone and a can of soda. Assume the ice cream and the cone are solid (no air gaps inside anywhere), and that the cone is a regular cone, with the portion of ice cream sticking up from the top of the cone forming a perfect half sphere. Let's also assume for simplicity that the soda can is a perfect cylinder. The ice cream scoop and the top of the cone both have a diameter of 12 cm, and the height of the cone (NOT including the ice cream!) is 15 cm. If the soda can has a diameter of 9 cm., how tall would the soda can need to be in order to contain the same volume as the cone with the ice cream?

This is my first post in this club and it is also my last!

What Does Math Best Relate To -> A) Computer Programming B) Artificial Intelligence C) Instruments D) Other (Write in Comments) Post a Letter, and you can explain why if you would like. P.S. If you pick "D", then don't say chess. That's obvious lol

This is the last lottery until May. Put a smiley face like this: to participate in a pi lottery of 30 pi. You can put any smiley you want, and you can put 1-5 smileys. Let's start! P.S. The lottery registration ends on April 27. Lottery Results will be posted soon after that.

https://www.chess.com/news/view/admin-coordinator-spots Please read the entire thing. Inviting is necessary.

Join this: https://www.chess.com/club/matches/math-and-chess-club/1116122 Thank you!

It’s all about that function f(n), shown above, which takes even numbers and cuts them in half, while odd numbers get tripled and then added to 1. Take any natural number, apply f, then apply f again and again. You eventually land on 1, for every number we’ve ever checked. The Conjecture is that this is true for all natural numbers. P.S. This problem has never been fully solved, but a great attempt was made by: Terence Tao. Mathematicians are so close! https://www.popularmechanics.com/science/math/a29033918/math-riddle-collatz-conjecture/

What is the name of this? Winner gets 5 pi.

Put a smiley face like this: to participate in a pi lottery of 30 pi. You can put any smiley you want, and you can put 1-5 smileys. Let's start! P.S. The lottery registration ends April 2. Lottery Results will be posted soon after that.

Rules: 1. Don't copy and paste from the internet. 2. The riddle actually needs an answer. The Riddles Are 3 pi each if they are made by yourself.

I just learned about martingales and the martingale gambling strategy, so I'm no expert on the topic, but I'll summarize it the best I can (with the help of Wikipedia, of course!). https://en.wikipedia.org/wiki/Martingale_(probability_theory) A martingale is a situation where the conditional expectancy (i.e. average expected value) is equal to the current / most recent value. For example, in a random walk (e.g. taking n steps in random directions starting at the origin of Cartesian xy graph), the conditioned expectancy is the starting position. I am wondering: does chess ELO counts as a martingale? My gut tells me that it does, but I don't understand martingales or ELO well enough to say for sure.

https://www.piday.org/pi-quiz/ How many can you get? Also, this might take a long time to load on a slow device, but here are the first million digits of pi. https://www.piday.org/million/ Good luck!

1. I can only live where there is light, but I die if the light shines on me. What am I? 2. A man wants to enter an exclusive club, but he doesn't know the password. Another man walks to the door and the doorman says 12, the man says 6, and is let in. Another man walks up and the doorman says 6, the man says 3, and is let in. Thinking he had heard enough, he walks up to the door and the doorman says 10, he says 5, and he isn't let in. What should he have said? 3. You want to boil a two-minute egg. If you only have a three-minute timer (hourglass), a four-minute timer and a five-minute timer can you boil the egg for only two minutes?

I have a calculator that can display ten digits. How many different ten-digit numbers can I type using just the 0-9 keys once each, and moving from one keypress to the next using the knight’s move in chess? (In chess, the knight move in an L-shape – one square up and two across, two squares down and one across, two squares up and one across, and other like combinations)

Let's recite the numbers of pi! Rules: Only post one number. If the one before you isn't correct, fix it and you get 1 pi. Let's begin: 3.14..