Prove that the number $ (\underbrace{9999 \dots 99}_{2005}) ^{2009}$ can be obtained by erasing some digits of $ (\underbrace{9999 \dots 99}_{2008}) ^{2009}$ (both in decimal representation).
Yudi Satria, Jakarta
Outline: Write $ 999...9=10^n-1$. So $ 999...9^{2009}=(10^n-1)^{2009}=\sum\binom{n}{k}10^{nk}-\binom{n}{k-1}10^{n(k-1)}$. Therefore $ 999...9^{2009}$ can be written as blocks of numbers. It is easy to prove the each block of $ (\underbrace{9999\dots 99}_{2008})^{2009}$ can be erased to get a block of $ (\underbrace{9999\dots 99}_{2005})^{2009}$