6. If the tens digit of $a^{2}$ can be $1,3,5,7,9$, then the units digit of $a$ is ).
(A) must be 4
(B) must be 6
(C) can be 4 or 6
(D) $2,4,6,8$ are all possible.
6. C.

Let $a=10b+c$, where $c$ is taken from $0,1,2, \cdots, 9$. Write $c^{2}$ as a two-digit number: $00,01,04,09,16,25,36,49,64,81$. Among these, only $c=4,6$ have an odd tens digit.
Also, $a^{2}=(10b+c)^{2}$
$$
=2 \times (5b^{2}+bc) \times 10 + c^{2}.
$$

It can be seen that the tens digit of $a^{2}$ is an even number plus the tens digit of $c^{2}$.
When the tens digit of $a^{2}$ is an odd number $1,3,5,7,9$, the units digit of $a$ can only be 4 or 6.
When the units digit of $a$ is 4 or 6, we have
$$
4^{2}=16, 14^{2}=196, 24^{2}=576, 34^{2}=1156,
$$
$44^{2}=1936$, whose tens digit covers $1,3,5,7,9$;
$$
6^{2}=36, 16^{2}=256, 26^{2}=676, 36^{2}=1296,
$$
$46^{2}=2116$, whose tens digit covers $1,3,5,7,9$.