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# Submitting Problems to Brilliant is going away for a few months

Over the past several months, we have received an impressive number of challenging, clever, and fun problems from all of you. Writing good problems is hard, and we have been inspired by the number of you who are willing and able to produce great ones.

The problems you all have submitted have taught us that only a small portion of problems that people write, are problems that fit our very strict criteria to be featured in our problem sets. What they have also taught us, is that our strict criteria excludes tons of interesting problems that all of you would find fun to solve. Further more, if a math problem is unclear, or needs work; it would be much better to have the community provide detailed constructive feedback on the problem, rather than be dismissed by us.

Over the next several months we will be working to reinvent the way problems and discussions are published and disseminated on Brilliant. We would like our members to have the ability to publish problems, hosted in our interface, and disseminate them to the people they like solving problems with on Brilliant. Unfortunately, this will take awhile to build.

We have taken down the problem submitting button, to divert our efforts toward enabling problem publishing for everyone. The time saved by not having to maintain the currently inefficient process of problem submission, will be better applied toward making the site experience better for everyone.

In the mean time... if you have a good problem that you think others here would enjoy, post it to discussions. If they are disciplined, your audience can avoid looking at people's answers until they have tried to solve it. While not the same, it somewhat approximates the experience of our weekly problems.

Cheers,

The team behind Brilliant.org

Note by Peter Taylor
3 years, 10 months ago

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I really salute the people like Peter working for this website. They are very enthusiastic and trying to make the site better and better. My best wishes! · 3 years, 10 months ago

Might be helpful to have an option that thread authors could toggle that hides replies until the reader pushes a "Show" button, say. That way, there's no chance of having part of the solution onscreen.

Yes, there's always finagling we can do client-side to hide it (resizing window, using a second window to cover the partial solution, etc.), but once part of the solution has been on the screen and you've seen it, even peripherally, you can never be sure how much of that solution registered in your brain before you blocked it out. Even accidentally recognizing a particular function or number can be a give-away.

I know that's yet more work for you guys, but....

(Could also have a separate "User-Submitted Problems" subforum where this feature is always on for all posts. In fact, that may not be such a bad idea even without the solution-hiding feature.) · 3 years, 10 months ago

Sounds like a good idea and it would definitely be doable. I would say they could implement such a feature from a day to maybe a week depending on how complex their forum infrastructure is. · 3 years, 10 months ago

Yeah, a spoiler tag would be helpful. · 3 years, 10 months ago

How Do I Submit issues now? · 3 years, 9 months ago

There are a set of questions from RMO-1990....The questions are as follows:

1) N is 50 digit number. All the digits except 26th digit from the left are 1 . If N is divisible by 13 ,then find the 26th digit?

2)A square sheet paper ABCD is folded such that B falls on the midpoint M of CD . Then the crease divides the BC in the ratio a/b , where a and b are coprime integers. Then * a+b = ? * · 3 years, 10 months ago

well...i don't know how to use typo... · 3 years, 10 months ago

I have one.

What is the remainder when $$2^{2013}$$ is divided by $$5$$ · 3 years, 10 months ago

Since $$2^2\equiv -1\pmod{5}$$, $2^{2013}=2(2^{2012})\equiv 2(-1)^{1006}\equiv\boxed{2}\pmod{5}$ Alternatively, use FLT or find the pattern. · 3 years, 10 months ago

Daniel C. - What's FLT ? · 3 years, 10 months ago

Fermat's Little Theorem · 3 years, 10 months ago

The last digit of $$2^{2013}$$ is 2 so the remainder is 2. · 3 years, 10 months ago

Wow, crazy simple answer. · 3 years, 10 months ago