, 2021

Solution:
We require that $n \geq 15$ so that the product can be divisible by $25$ without being even. In addition, for any $n > 15$, if we can acquire all residues relatively prime to $100$, we may multiply them by some product of $\{1,2,4,5,15\}$ to achieve all residues modulo $100$, so it suffices to acquire only those residues. For $n = 15$, we have the numbers $\{3,7,9,11,13\}$ to work with (as $1$ is superfluous); these give only $2^{5} = 32$ distinct products, so they cannot be sufficient. So, we must have $n \geq 17$, whence we have the numbers $\{3,7,9,11,13,17\}$. These generators are in fact sufficient. The following calculations are motivated by knowledge of factorizations of some small numbers, as well as careful consideration of which sets of numbers we have and haven't used. It is also possible to simply write out a table of which residues relatively prime to $100$ are included once each number is added, which likely involves fewer calculations.

First, consider the set $\{3,11,13,17\}$. This set generates, among other numbers, those in $\{1,11,21,31,51,61\}$. Since $\{7,9\}$ generates $\{1,7,9,63\}$, which spans every residue class mod $10$ relatively prime to $10$, we need only worry about
$$\{41,71,81,91\} \times \{1,7,9,63\}$$
Since $41$ can be generated as $3 \cdot 7 \cdot 13 \cdot 17$ and $91$ can be generated as $7 \cdot 13$, we need not worry about these times $1$ and $9$, and we may verify
$$41 \cdot 7 \equiv 87 \equiv 11 \cdot 17, \quad 91 \cdot 63 \equiv 33 \equiv 3 \cdot 11$$
and
$$91 \cdot 7 \equiv 37 \equiv 3 \cdot 9 \cdot 11 \cdot 13 \cdot 17$$
using the method we used to generate $49$ earlier. So, we only need to worry about
$$\{71,81\} \times \{1,7,9,63\}$$
We calculate
$$71 \equiv 7 \cdot 9 \cdot 17, \quad 71 \cdot 9 \equiv 39 \equiv 3 \cdot 13, \quad 71 \cdot 63 \equiv 73 \equiv 3 \cdot 7 \cdot 13$$
each of which doesn't use $11$, allowing us to get all of
$$\{71,81\} \times \{1,9,63\}$$
so we are only missing $71 \cdot 7 \equiv 97$ and $81 \cdot 7 \equiv 67$. We find
$$97 \equiv 3 \cdot 9 \cdot 11$$
and
$$67 \equiv 3 \cdot 9 \cdot 13 \cdot 17$$
so all numbers are achievable and we are done.