【Question 4】
There is a repeating decimal $0.2 \dot{5} 8 \dot{7}$. The sum of the digits in the 1st, 99th, 199th, and 299th positions of its decimal part is $\qquad$
【Analysis and Solution】Cycle problem.
$0.2 \dot{5} 8 \dot{7}$, the 1st digit after the decimal point is 2;
$(99-1) \div 3=32 \cdots \cdots 2,0.2 \dot{5} 8 \dot{7}$, the 99th digit after the decimal point is 8;
$(199-1) \div 3=66,0.2 \dot{5} 8 \dot{7}$, the 199th digit after the decimal point is 7;
$(299-1) \div 3=99 \cdots \cdots 1,0.258 \dot{7}$, the 299th digit after the decimal point is 5;
The sum of the digits in the 1st, 99th, 199th, and 299th positions of the decimal part of the repeating decimal $0.2 \dot{5} 8 \dot{7}$ is $2+8+7+5=22$.