G6.1 $2^{a} \cdot 9^{b}$ is a four digit number and its thousands digit is 2 , its hundreds digit is $a$, its tens digit is 9 and its units digit is $b$, find $a, b$.
$$
2^{a} \cdot 9^{b}=2000+100 a+90+b
$$

If $a=0,9^{b}=2090+b$
$$
9^{3}=729,9^{4}=6561
$$
$\Rightarrow$ No solution for $a$
$\therefore a>0$ and $0 \leq b \leq 3,2000+100 a+90+b$ is divisible by 2 $b=0$ or 2
If $b=0,2^{a}=2090+100 a$
$$
2^{10}=1024,2^{11}=2048,2^{12}=4096 \text { and } 0 \leq a \leq 9
$$
$\Rightarrow$ No solution for $a$
$\therefore b=2,2000+100 a+92$ is divisible by 9
$2+a+9+2=9 m$, where $m$ is a positive integer
$$
a=5, b=2
$$

Check: $2^{5} \cdot 9^{2}=32 \times 81=2592=2000+100(5)+90+2$