Prove that for any integer $k$ ($k \ge 2$) there exists a power of $2$ that among its last $k$ digits, the nines constitute no less than half. For example, for $k = 2$ and $k = 3$ we have the powers $2^{12} = ... 96$ and $2^{53} = ... 992$.

HIDE: original wording Probar que para cualquier k entero existe una potencia de 2 que entre sus ultimos k dıgitos, los nueves constituyen no menos de la mitad.