I have a basic understanding of math (mostly what you are taught until you're 18 in basic math class), but enjoy attempting to solve hard problems where you use logic and basic math.

For example I remember a problem in a national math competetion that was:
25 knights of the round table sit in seats beside each other (round table).
3 of them draw lots randomly, to see who will slay the newest bothersome dragon.
What is the probability that at least 2 of the 3 knights are sitting next to eachother?
-> I got this @16 y/o, remember it being a fun one.

or:
the number: 2^2000-1 ends with what number?
-> I remember not solving this in the time I were alloted, though the logic of solving it is quite easy when you realize the pattern

or geometry problems etc

These problems where solved without the use of a calculator. Just pen and paper

I'd love to get back into doing these kind of problems, and challenging my brain (I've probably forgotten a lot of the math rules though)


So my question is:
Does anyone know websites with problems that I describe?
Or a link to a set of problems from previous math competetions/tests where calculator and aids weren't allowed.
The answers doesn't nescessarily have to be available.

This post was edited by Dragon_Reborn on Oct 16 2015 06:36am

Perhaps https://projecteuler.net/ has what you are looking for. Many of them require logic from a computer to solve. but it is a good learning curve for getting the computer to help you solve problems of this nature.

This post was edited by TheStealthTarget on Oct 16 2015 07:44am

look into intro to number theory or google for high school math competitions.

So would someone solve the probability one - Knights of the round table? I always get confused with permutations and combinations. Obviously, 3 random knights would be 3/25. However, any 2 together we would have to start subtracting. Help?

This site looks very, very interesting. Thanks for posting.
At first I thought I would start with problems where the use of a calculator/computer aren't needed, but the description on the site fits well with what I'm looking for.

I just hope I won't get stuck because I lack education in mathematical terms and methods.
I bookmarked it and will look into it more at a later time.


@carteblanche
I'll google number theory and read up on what that entails. I did some searching and found a nice website with problems for high school math competetion in my country (abel konkurransen). I actually entered all 3 years in high school, and won all of them for my school. Kind of funny because the 1st year I only got like 9/20 questions right - I found this exact test on the website I remembered, and that is where the 25 knights problem was from. Some teachers got mad at me for not participating in the 2nd rounds though, but then I'd have to drop gym classes etc, kind of regret it now.
I'll try to do them again, and see how I do, probably a lot of hard questions to keep my occupied for a while.


@joeinfhills
This problem is actually rather tricky, as it's easy to mess up and forget to account for certain scenarios.
I've asked this one to someone studying math in college, but they didn't get it right either.
I did this again now, and posting answer for you here:

(P) = p(2A1) + p(3A1,2 - variation 1) + p((3A1,2 - variation 2)
P = probability of at least 2 knights sitting next to eachother
A refers to adjacent. for example 2A1: knight nr 2 is next to nr 1. 3A1,2: knight nr 3 is next to either nr 1 or nr 2.
variation 1: this is when the 2nd knight is not seated next to the first, but with 1 seat in between (only 3 seats where nr 3 can sit to be adjacent)
variation 2: this is when the 2nd knight is not seated next to the first, but with at least 2 seats in between (4 seats where nr 3 can sit to be adjacent)

(P) = 2/24 + (2/24)*(3/23) + (20/24)*(4/23)
(P) = 132/(24*23)
(P) = 11/46

I don't know of a particular site(though you can probably easily google them) but I have a bunch of problems from past worksheet/contest that I've done. Here's a couple of questions you can try out

1. Solve for x. Note figure not drawn to scale.


2. Suppose there are n people in a duel. They shoot at the same time. Each person shoots at one other person at random(cannot shoot themselves). Assume that each person can hit their target 100% of the time and if someone gets shot, they die. What is the probability that at least one person survives?


Let me know if these are too easy/hard or just right difficulty, because I have a bunch of these types of problems if you'd like to see more.

I've never felt so dumb until I clicked in here. I came here for shop math help I feel like I'm going to get made fun of here

Well... I certainly was not expecting the answer to be greater than 3/25. So it must be that I don't understand the question. What does "3 of them draw lots randomly, to see who will slay the newest bothersome dragon" mean?

I had assumed 3 out of 25 were randomly picked and was was the probability that at least 2 of them were beside each other.

[QUOTE=cdexswzaq,Oct 20 2015 09:47pm

2. Suppose there are n people in a duel. They shoot at the same time. Each person shoots at one other person at random(cannot shoot themselves). Assume that each person can hit their target 100% of the time and if someone gets shot, they die. What is the probability that at least one person survives?

[/QUOTE]

I am having a hard time following the question. "Each person shoots at one other person at random", means to me that 2 people cannot target 1 person. In this case, the only way to guarantee "n" people target "n" people is for them to stand in a circle. If that's the case, they all always die.

It should be interpreted as each person can choose to shoot at any of the other (n-1) people; otherwise like you said it is trivial and they will all always die

This post was edited by cdexswzaq on Oct 21 2015 11:07am