【Question 2】
Move the decimal point of number A one place to the right to get number B, and move the decimal point of number A two places to the left to get number C. It is known that the sum of A, B, and C is 181.665, and number A equals $\qquad$.
【Analysis and Solution】The problem of multiple times.
(Method One)
Moving the decimal point of number A one place to the right gets number B, then B is 10 times A;
Moving the decimal point of number A two places to the left gets number C, then A is 100 times C;
C equals $181.665 \div(1+100+100 \times 10)=0.165$;
A equals $0.165 \times 100=16.5$.
(Method Two)
Let A be $x$;
Moving the decimal point of A one place to the right gets B, then B is 10 times A, B equals $10 x$;
Moving the decimal point of A two places to the left gets C, then C is 0.01 times A, C equals $0.01 x$;
$x+10 x+0.01 x=181.665$; solving for $x=16.5$; A equals 16.5.