1. Find the sum of the digits of the number $A$, if $A=2^{63} \cdot 4^{25} \cdot 5^{106}-2^{22} \cdot 4^{44} \cdot 5^{105}-1$. Answer: 959.
Solution. Since $2^{63} \cdot 4^{25} \cdot 5^{106}=2^{113} \cdot 5^{106}=2^{7} \cdot 10^{106}$ and $2^{22} \cdot 4^{44} \cdot 5^{105}=$ $2^{110} \cdot 5^{105}=2^{5} \cdot 10^{105}$, then $A+1=2^{7} \cdot 10^{106}-2^{5} \cdot 10^{105}=\left(2^{7} \cdot 10-2^{5}\right) \cdot 10^{105}=$ $1248 \cdot 10^{105}$. Therefore, the number $A$ is $1248 \cdot 10^{105}-1=1247 \underbrace{999 \ldots 999}_{105 \text { digits }}$, and the sum of its digits is $1+2+4+7+9 \cdot 105=959$.