All initial $1$-digit factors must clearly be odd, or they are in $\{3,5,7,9\}$, since we can disregard the effect of any factor $1$. Moreover, since $9=3^2$, $n=3^a\cdot 7^b\cdot 5^c$ for some non-negative integers $a,b,c$. Assume that $n$ does not end in $5$, or $n=3^a\cdot7^b$ for some non-negative integers $a,b$. Assuming that $1$-digit numbers have a $0$ tens digit (and hence even), we show by induction on $a+b$ that the tens digit in $n$ would always be even. If $a+b=1$ this is clearly true since $n$ would be a $1$-digit number. Now, let $n=3^a\cdot7^b$ such that the proposed result is true for $a+b$, and consider $3n$ and $7n$. Note that $n$ can be written as $10u+v$, where $v\in\{1,3,7,9\}$ because $n$ is odd and has no $5$ factor, and where $u$ is even by hypothesis of induction. Thus, the tens digit of $3n$ is the sum of the units digit of $3u$ (even because $u$ is even) and the tens digit of $3v$. But $3v\in\{3,9,21,27\}$ has always an even tens digit. Similarly, $7v\in\{7,21,49,63\}$ also has an even tens digit, and the result is also true for $a+b+1$. We therefore reach a contraction, and at least one factor of $5$ must appear, hence $n$ ends in $5$.

Claim: The tens digit of powers of $3$ are always even. Proof. Modulo $20$, we have $3^{4k} \equiv 1$, $3^{4k+1} \equiv 3$, $3^{4k+2} \equiv 9$, $3^{4k+3} \equiv 7$. $\blacksquare$ The answer is $5$ only, achieved by $5 \cdot 7 = 35$. If $n$ is the product of one-digit numbers, has all odd digits, and does not end in $5$, then $n$ is odd and $n = 3^a 7^b$ for some nonnegative integers $a$ and $b$. However, $7 \equiv 3^3 \pmod{20}$, so $n \equiv 3^{a+3b} \pmod{20}$ and the claim now kills the problem.

We claim the unit digit can only be five, which works from $n=3\cdot 5.$ If the units digit is not five, $n$ must be of the form $3^a\cdot 7^b$ as any factors of $5$ or $2$ would make the units digit $\{0,5\}$ or $\{0,2,4,6,8\},$ respectively. Notice \[3^a\cdot 7^b\equiv 21^b\cdot 3^{a-b}\equiv 3^{a-b}\pmod{20}.\]Notice $\operatorname{ord}_{20}3=4$ and $3^0\equiv 1,$ $3^1\equiv 3,$ $3^2\equiv 9,$ and $3^3\equiv 7\pmod{20}.$ Hence, $n\equiv 1,3,9,7\pmod{20}.$ If the tens digit of $n$ had been odd, then the remainder when $n$ is divided by $20$ would have a tens digit, a contradiction. Hence, $n$ has an even tens digit and the units digit cannot be $1,$ $3,$ $7,$ or $9.$ $\square$

The answer is 5 only; for a construction, $15=3 \cdot 5$. Now, in general I claim that if $n$ only contains prime factors $3$ and $7$, then the tens digit of $n$ is even. This can simply be done by finite case checking mod 20, as $3^4 \equiv 1 \pmod {20}$ and $7^4 \equiv 1 \pmod {20}$ so there are only $16$ cases. Remark. There is a nice trick mentioned above where we kill the $7$'s, but why think when you can bash?

How do you see that mod 20 is the right way to go?

math12345678 wrote: How do you see that mod 20 is the right way to go? $\pmod{20}$ gives you the information about the parity of the tens digit. You basically want to prove that if it doesn't end in a $5$, the tens digit is even. After some trying out ($3 \cdot 7$, $9 \cdot 7$), you find a pattern and see that the $10$s digit is even. By using $\pmod{20}$ you can show this and complete the problem.

e claim that the answer is 5 only. Define the body of a positive integer $n\geq10$ as $\lfloor n/10 \rfloor$ (essentially cutting off the units digit). Note that all odd numbers that are expressible in that way are of the form $$3^a5^b7^c.$$ Claim: For all nonnegative integers $(a,c)$, the body of $3^a7^c$ is even. We will use induction. $(0,0)$ obviously satisfies this. Now, note that all numbers of this form end in either 1, 3, 7, or 9. If the units digit is 1, multiplying by 3 makes the body go from $B$ to $3B$, and multiplying by 7 makes the body go from $B$ to $7B$. If the units digit is 3, 3x makes $B$ go to $3B$ and 7x makes $B$ go to $7B+2$. If the units digit is 7, 3x makes $B$ go to $3B+2$ and 7x makes $B$ go to $7B+4$, and if the units digit is 9, 3x makes $B$ go to $3B+2$ and 7x makes $B$ go to $7B+6.$ We can see that in all of these cases, the body remains even, so we have shown the claim by induction. Therefore, if there are no factors of 5, the body must be even, and the number will therefore have an even digit. Therefore, the answer is 5 only, which can be achieved by 25, so we are done.

wait oops lol i overcomplicated this "the body is even" is just mod 20 is between 0 and 9 which you can easily just bash

The answer is only $5$. For the construction, $5\times 5 \times 3 = 75$ works. Now FTSOC assume that the unit digit is not $5$. Thus $n$ must be of the form $3^a\cdot 7^b = 3^i \cdot 7^j \cdot (21)^k$ where $0\le i,j\le 1$ and $a,b\ge 1$. I claim that the ten's digit is always even. Then note that the remainders that $21^k$ leaves $\pmod{100}$ are $S = \left\{01,21,41,61,81\right\}$. Then we can check that $3 \times S$ and $7 \times S$ all have their ten's digit even. Apart from these, we only need to check the prime powers. Note that $3^5 \equiv 43 \equiv 7^3 \pmod{100}$. This means that $3^k$ with $k\ge 5$ can be expressed as $3^i \cdot 7^j \cdot (21)^k \pmod{100}$. Thus we just need to check for $k\le 4$. Similarly, we need to check $7^l$ for $l\le 2$. The cases $\left\{3^3,3^4,7^2\right\}$ are easy to check by hand and the others do not work since they are $<10$. Thus we are done.

The answer is $\boxed{5}$, which is clearly attainable. The units digit is odd, so the number is formed by multiplying some combination of $3$s, $5$s, $7$s, and $9$s. Suppose that the units digit of $n$ is not $5$. Then, $n\equiv 3^k\pmod {20}$ for some integer $k$ ($3^3\equiv 7\pmod {20}$). Since powers of $3$ only have even tens digits, this is impossible.