$30 \cdot 30$ has exactly 35 consecutive natural numbers whose integer parts of the arithmetic square roots are the same. Then this same integer is
(A) 17.
(B) 18.
(C) 35.
(D) 36.
(China Junior High School Mathematics League, 1990)
[Solution] Let the smallest of these 35 consecutive natural numbers be $n^{2}$, and the largest be $(n+1)^{2}-1$. Since $(n+1)^{2}-n^{2}=35$. That is, $2 n+1=35$. Thus, $n=17$. Therefore, the answer is $(A)$.