I support you as always, you're doing the right thing!

still gonna post puzzles here? It's been days and I'm eager to solve more.

not everyone is a native English speaker, I believe what he meant was a civics class. @ismynewmail.bsky.social

Thank you for sharing this, I'm also going to give a donation

Wspaniały Wiersz!

1, Burgundy
2, Amber
3, Sapphire

now I'll name each given information as A, B, C, D(top-bottom)
D, thus 0 isn't correct
C, thus 9, 7 are correct
B, thus 7 is the very correct one
thus 2 isn't correct
and D, thus 6 isn't correct
A, thus 3 is correct
it's placed wrong, thus 3 is the first digit
thus 3, 7, 9 is the solution
QED

(Part III)
ą^|g1*g2*...gz|=(p1^r1)(p2^r2)...(pz^rz), where r1...rz∈Z, r1...rz≠0
now time both sides of the equation with all possible pi^|ri| st ri(i=1...z) is a negative number.
then the left side is a composite number with ą as its factor; the right side is not. Thus there's contradiction.
QED

(Part II)
let c1=f1/g1, c2=f2/g2... where f1, f2...∈N+ and g1, g2...∈Z(none of them are 0)
now that the amount of pi(i=1,2..z) is finite, there exists a prime number different from them all, let's use a symbol ą for it.
Assume ą=(a1^x1)(a2^x2)...=(p1^(f1/g1))(p2^(f2/g2))...(pz^(fz/gz))

I mean "not all *positive* rational numbers can..." by the way