tag:blogger.com,1999:blog-444408592833092365.post3360104783671552421..comments2023-06-29T04:02:58.043-06:00Comments on México rumbo a la IMO: Problema del diaDavid (sirio11)http://www.blogger.com/profile/13765612869477578855noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-444408592833092365.post-50977190664624957812011-06-14T16:58:21.849-05:002011-06-14T16:58:21.849-05:00Para cada $r\in \mathbb{N}$, sea $n=k+1$ y $m=r(k+...Para cada $r\in \mathbb{N}$, sea $n=k+1$ y $m=r(k+1)!-1$. Esas son infinitas parejas, y cada una funcióna porque <br />$$\frac{(m+n-k)!}{m!n!}=\frac{(m+1)!}{m!(k+1)!}=\frac{m+1}{(k+1)!}=\frac{r(k+1)!}{(k+1)!}=r\in\mathbb{N}$$<br />QEDAnonymoushttps://www.blogger.com/profile/14839425512917090298noreply@blogger.com