True or False?

We proceed by induction. For $K = 1,$ we see that $N = 5$ satisfies the given requirements. Now, suppose that the $n$ -digit number $N = \overline{a_na_{n - 1} \dots a_2a_1}$ is divisible by $5^n.$ We will prove that there exists an odd digit $a_{n + 1}$ such that $\overline{a_{n + 1}a_na_{n - 1} \dots a_2a_1}$ is divisible by $5^{n + 1}.$

Since $5^n|\overline{a_na_{n - 1} \dots a_2a_1},$ we can write $\overline{a_na_{n - 1} \dots a_2a_1}$ in the form $5^nk,$ where $k$ is some positive integer. Next, we see that

$\begin{aligned} \overline{a_{n + 1}a_na_{n - 1} \dots a_2a_1} &= 10^na_{n + 1} + \overline{a_na_{n - 1} \dots a_2a_1} \\ &= 10^na_{n + 1} + 5^nk \\ &= 5^n(2^na_{n + 1} + k). \end{aligned}$

Our problem, then, reduces to finding an odd digit $a_{n + 1}$ such that $2^na_{n + 1} + k \equiv 0 \! \pmod{5}.$ Noting that $2^n \equiv \pm 1, \pm 2 \! \pmod{5},$ we can reduce this congruence to $\pm a_{n + 1} \equiv -k \! \pmod{5}$ or $\pm 2a_{n + 1} \equiv -k \! \pmod{5}.$ In all four cases, we can see that there exists a non-negative integer $i$ less than 5 such that $a_{n + 1} \equiv i \! \pmod{5}.$ (For the -1 case, we merely divide by -1. For the $\pm 2$ cases, we manipulate the RHS of the congruence until it is even, then divide both sides by $\pm 2,$ which preserves the modulo since $\gcd(2, 5) = 1.$ )

If $i$ is odd, we take $a_{n + 1} = i,$ which will make $\overline{a_{n + 1}a_na_{n - 1} \dots a_2a_1}$ divisible by $5^{n + 1}.$ If $i$ is even, we take $a_{n + 1} = i + 5,$ which will achieve the same result.

Our inductive step is complete, so for all positive integers $K,$ there exists an $N$ with $K$ odd digits such that $5^K|N. \, \, \blacksquare$