Hey, this forum game is based on Yosh's idea of a game, where you solve the mathematical or logical problem above and post a new problem for the next player to solve etc. How to Play (i) I'll start with the first problem (ii) Now it's your turn. You try to solve the current problem as best you can and write your answer. (iii) After that, you formulate the next mathematical or logical problem for the next player. Be as creative as possible Rules* (i) Have fun (ii) Limit the problems to mathematical and logical tasks (iii) The player that posted the problem has to confirm your solution before you can post a new tast (iv) Don't create unsolvable tasks (v) No math homework Let's get started Krysyy and Aikar invited some staff to dinner and received the following feedback: (i) If Eviltoade does not come, JDHallows does not want to come either; (ii) At least one of BurgerKnight and Luckygreenbird comes; (iii) Either Eviltoade or chickeneer comes, but not both; (iv) Either chickeneer and Luckygreenbird come, or both do not; (v) When BurgerKnight comes, Luckygreenbird and JDHallows come too. Who's coming to eat now? *might change in the future according to feedback of the players
I guess if gaming says so..... Lets do this equations like (it's the best game ever) You have the numbers: 1, 0, 7, 10, 1, 4 And the signs: -, -, -, ÷, ÷ The goal is to write an equation, keep pemdas in mind, that gets you 51. You can only use each sign and number once, but you can use parentheses as much as you like! Have fun
Guess I was wrong, I was thinking the same thing tbh, let's see again...it has to be toade who comes since he can't go if the Chicken goes and if JD comes then Toade must be there as stated in the first statement. It wouldn't make sense if both Chickeneers and Toade went as stated in the 3rd statement. There must be something I'm missing
But when lucky comes, chicken also has to come (iv). Your approach is the right one. Try to assume that the staff member X comes and semantically deduce who must also come and who can not. As soon as you find a contradiction, your assumption is wrong and X certainly does not come. After that, you can try another one.
Is it Luckygreenbird and Eviltoade? And, because there isn't something that describes whether he comes or not in this case, JDHallows does or does not come based on whether he feels like it.
Thank you for playing, Tom. But also your solution is not quite right When Luckygreenbird comes, chickeneer also has to come (iv) But when Eviltoade comes, chickeneer can not come (iii) If I made no mistake, the solution is also unique and has no free variables.
Whoops, for some reason I kept switching those two around, and after scrolling back up and down at least 5 times I still put the wrong name. What I meant to say: Is it Luckygreenbird and chickeneer?
Congratz! We have a winner Yeah, I also had to scroll a lot to put in the right names to answer you ^^ I like your outside the box solution very much xD
I'll add the rule, that the solution has to be verified by the player who posted it before a new problem is posted. If it's okay with Tom, we can continue with Yosh's post
Let me post a few problems. Of all numbers with five digits there are a where the product of those digits is equal to 25 and b where the product of the digits is equal to 15. What is a/b equal to? What is the smallest possitive whole number that only exists of the digits 2, 4 and 8, of which each are used at least two times, and the number itself isn't deviseble by 4? If we write all the numbers between from 1 to 30 000 in one line of digits (12345678910111113...30000), how many times will the number 2013 show up? Twenty studens made a test, no two studens made the same amount of mistakes. No question was answered correctly by more than three people. What is the smallest amount of questions this test could have had? ---------- And now a very dificult one, for the people studing maths at Universety, or are just really good at maths: Choosing n+1 points on an n-dimentional sphere. Proof that the chance the body enclosed by the ribs between those points to contain the center of the sphere expressed in n, with n > 1 equals 2^-n
One at a time work 4 ya? We're currently doing this one: Lets do this equations like (it's the best game ever) You have the numbers: 1, 0, 7, 10, 1, 4 And the signs: -, -, -, ÷, ÷ The goal is to write an equation, keep pemdas in mind, that gets you 51. You can only use each sign and number once, but you can use parentheses as much as you like! Have fun But after ward we can do yours!
I'm making a joke, I'm a maths student and thease are random questions from the national Dutch Maths championchips, exept from the last one, the last one /is/ a test I made on universety. Yes, the test was just that question, and we had four hours to solve it (and I did not finish it)
That's terrible. D: (in the traditional sense of the word, not the meaning of 'rubbish') I'm sorry, I have been studying for a Maths exam that I need to 'resit' (I wasn't able to attend the original exam), and I'd rather not exert myself on Maths in my spare time right now.
