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I do not think I will be able to produce a very mathematical solution, sorry.

I liked this problem. Post more of these!

Okay. Do you know why you revived this thread?

Why do you want a more "mathematical" solution?

See the Brilliant link and click on Reveal Solutions

So this is a problem from USAJMO?

Yes, but if you think it is very far away from mathematics, you're wrong.

Many interesting Combinatorial problems requires computational power for solving. But does that mean computers can reduce the problems into pure bashing? Nope! As you can see there, the backtracking solution can only solve upto the 10 by 10 chessboard whereas the one supplemented by enough research can solve upto a 174 by 174 chessboard.

I found another solution (much like this one) on another forum: https://brilliant.org/problems/too-many … hessboard/

If a < b (mod 2012) shouldn't ka < kb (mod 2012)?

You will need a computer for this: http://codegolf.stackexchange.com/quest … hess-board

One solution is a backtrack solution. Another guy (elsewhere) gave me a gf solution which I did not really understand that well.

I have reasons to believe the answer is 6148.

@phanthanhtom, Could you please look into this?

No, you have got long grasses in your backyard. Also. You have got bottles of soy sauce

Just use the large text box.

If you had a phone you would not need to buy a car, you could've just called the people you want to visit

That is really a good thing about Alice.

Here is my kaboobly doo machine: http://kabooblydoo.appspot.com

It works on the principle of Predictive text

Is Alice a member here?

Predictive texting is fun. Have you seen my Kaboobly Doo machine?

Nope. He proved the other thing too.

When f(n) is not of that form, the odds pair up

We are using strong induction here.

First we show that it is true for some of the beginning values by computation.

Then, we assume that for some value n, all of the preceeding values show the proposed property.

We now need to show that n also shows the same property, to complete the induction.

Note that f(n) = f(n-2^0) + f(n-2^1) + ... + f(n-2^k) + ...

We're interested in if any of these are odd. By the induction hypothesis, if there is such a term, it should be of the form 2^j - 1. This is because we've chosen to accept that only the output of such numbers are odd.

So, n - 2^k = 2^j - 1

This is a combinatorial search problem.

I think we should start writing a backtrack solver. I do not know how much time this method would take but at least we will discover a good many of the configurations

math9maniac wrote:

Hi Agnishom;

Tell me something. What relationship exists between Bob and Alice?

Bob can send encrypted messages to Alice. He does not find cellular networks secure.

I did not say proofs are Kaboobly Doo

Kaboobly Doo!

You did not have the proof. You did not even attempt it.

Here is the proof of the conjecture:

No matter how fast light is, the bottleneck effect counts.

What kind of points?

Over OEIS? Why should they care if some values are even?

Calvin gave me a proof