Easy: (13,13,10) = (8,15,17) EDIT: Wrong indeed, thank you, Paolo.

<t>Sashka,Lambert's bound is the following result. Consider a homogeneous linear equation in nonnegative integers a1 x1 + ... + an xn = 0 The set of solutions is closed under addition. There are solutions which can be represented as a nontrivial sum of other solutions and there are primitive solutio...

Yes, that's it!It is not even necessary for Tomania to have only one denomination. The bound is true if A is the highest denomination of Tomanian coins. I know this result as Lambert's bound from integer programming (for Jean-Luc Lambert).

<t>Two countries, Oisterlich and Tomania have common currency but different coinage. Oisterlich monetary system has m coins of denominations B1 < B2 < ... < Bm (due to high inflation rate B1 > 1).Tomania makes coins of single denomination A (what would you expect from a totalitarian state).Shopkeepe...

<t>Ok, ok, not only was there an error in my writing, but it was also incomprehensible altogether.And comparing it with Tomfr's solution, where given a number of the form 10^1999*A + B, where B has no zeros, 10^1999*A is subtracted off, the latter is certainly nicer.Tomfr, could you tell where does ...

Thank you, I have to correct this mistake. The intended definition of Z(x) is the position of the RIGHTmost zero counted from the RIGHT, starting from 0, so that Z(512) = 3, Z(4096) = 2,Z(11111101) = 1, Z(2^17) = 2 and Z(2^17 * 26) = Z(3407872) = 4.

<t>We shall fight them one by one!As Tomfr says, N is equal to 2^1999. Let K(x) be the number of digits in the decimal representation of x and let Z(x) be the position of the rightmost zero in the decimal representation of x counted from the right[*] starting from 0, so K(1024) = 4 and Z(1024) = 2, ...

<t>A more of an avian proof.Let's prove by induction on n that if n positive numbers sum up to Q, where n <= Q <= 2n - 2, then for any K, such that 1 <= K <= Q, there is a subsetthat sums up to K.For n = 1 the statement is trivially true, since 0 is not a positivenumber.To prove the induction step, ...

It is definitely more responsive than citeseer.ist.psu.edu and indexes stuff beyond mathematics/computer science.The only feature of citeseer that I miss is the citation contexts.

<t>This does not seem to be mentioned before.If the main load on the server is going to be C++ compilation, then having two or four processors is likely to improve performance almost proportionally to the number of processors. Such configuration may also pay back if the machine will run several nume...

<t>In refinement of the preceding post by AaronLet p, q and r be the proportions of UU, UR and RR individuals in the population respectively, then in the next generation we haver' = (r+ q/2)^2, p' = (p+q/2)^2 and q' = pq + rq + 2pr + q^2/2And certainly, p + q + r = 1. The set of solutions is an alge...