$28.816 \div(0.40+0.41+0.42+\cdots+0.59)$ The integer part of the value is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
[Solution] The original expression $>16 \div(0.6 \cdot 20)=16 \div 12>1$, and $\quad$ the original expression $<16 \div(0.4 \cdot 20)=16 \div 8=2$. Therefore, the integer part of the original expression is 1. Hence, the answer is $(A)$.