Aiscrim wrote: Let positive integers $a,\ b,\ c$ be pairwise coprime. Denote by $g(a, b, c)$ the maximum integer not representable in the form $xa+yb+zc$ with positive integral $x,\ y,\ z$. Prove that \[ g(a, b, c)\ge \sqrt{2abc}\] (M. Ivanov)

It is not true. Let $a<b<c$. Obviosly $g(a,b,c)\le ab-a-b<\sqrt{2abc}$ if $c>\frac{(a-1)(b-1)}{2}$. For example a=1 or a=2.

http://en.wikipedia.org/wiki/Coin_problem#n_.3D_3

Frobeneus number $g(a_1,...,a_n)$ for $(a_1,...,a_n)=1$ is defined as largest number N/ suth that not exist representation $N=\sum{}_x{}_i a_i, x_i\ge 0$. For n=2 $(a_1,a_2)=1$ $g(a_1,a_2)=a{}_1a{}_2-a_1-a_2$. Obviosly if $(a_1,...,a_k)=1, k<n$, then $g(a_1,...,a_k)\ge g(a_1,...,a_n)$. Therefore for $a<b<c, (a,b)=1$ $g(a,b,c)\le g(a,b)=ab-a-b$. When $c\ge (a-1)(b-1), (a,b)=1$ $g(a,b,c)=g(a,b)=ab-a-b<\sqrt{2abc}$.

People! this is not Frobenius' problem. Notice it is stated "... maximum integer not representable in the form $xa+yb+zc$ with positive integral $x,\ y,\ z$ ...". This makes a world of difference with Frobenius, where non-negative integers are allowed. Think for example to the more common and simpler case with two coprime positive integers $a,b$ only. Frobenius threshold is $(a-1)(b-1) - 1$. But try to represent $ab = xa + yb$, with integer $x,y > 0$. You need $a\mid y$ and $b\mid x$, thus $a\leq y$ and $b\leq x$, so $ab = xa + yb \geq 2ab$, absurd. Thus the threshold is at least $ab$ (and it can be easily shown it is precisely $ab$). Small differences in conditions may lead to largely different results.

mavropnevma wrote: People! this is not Frobenius' problem. Notice it is stated "... maximum integer not representable in the form $xa+yb+zc$ with positive integral $x,\ y,\ z$ ...". This makes a world of difference with Frobenius, where non-negative integers are allowed. Difference is small. If $g'(a,b,c)$ is largest number not representation in the form $xa+ya+zc$, then $g'(a,b,c)\le g(a,b,c)+a+b+c$.

any solution?

qwert159 wrote: any solution? Yes. A solution here: Consider how many pairs $(x, y)$ satisfy $xa+yb < \sqrt{2abc}$. This is clearly at most the number of lattice points with positive integer coordinates in the first quadrant, bounded by the line $xa+yb < \sqrt{2abc}$, which is a triangle. But the number of lattice points is less than the triangle area, which is $\frac{1}{2} \cdot \frac{\sqrt{2abc}}{a} \cdot \frac{\sqrt{2abc}}{b} = c$. Therefore, some $xa+yb$ will not hit all values $\pmod{c}$. Therefore, $xa+yb+zc$ can't hit every integer that is at least $\sqrt{2abc}$... since it won't hit everything $\pmod{c}$.

Anyone has the official statement?

R8kt wrote: Anyone has the official statement? Yes, I have. It coincide with #8.

micliva wrote: R8kt wrote: Anyone has the official statement? Yes, I have. It coincide with #8. where can I find official solutions to tuymaada 2002?

blackmetalmusic wrote: where can I find official solutions to tuymaada 2002? I think it is not for this topic, you can ask me in PM.