$29 \cdot 8$ in $\{1,2,3, \cdots, 99,100\}$ the number of $n$ that satisfies the condition that the tens digit of $n^{2}$ is odd is
(A) 10.
(B) 20.
(C) 30.
(D) 40.
(E) 50.
(45th American High School Mathematics Examination, 1994)
［Solution］The last two digits of a perfect square can only be the following possibilities:
$-00, -25, -$ odd $6, -$ even $1, -$ even $4, -$ even $9$

If the tens digit of $n^{2}$ is odd, it is only possible when the last digit is 6, which means the unit digit of $n$ is 4 or 6.
There are 20 such $n$'s between 1 and 100.
Therefore, the answer is $(B)$.