= Solution
\b[Yes.] A concrete example of <nonunique factorization in a congruence submonoid> is
$$
\boxed{21\cdot221=51\cdot91=4641.}
$$
The four factors are distinct and each is congruent to one modulo ten. The primes $3,7,13,17$ can be checked by trial division by two and three. Their elementary prime decompositions are
$$
21=3\cdot7,\quad221=13\cdot17,\quad51=3\cdot17,\quad91=7\cdot13.
$$
For any one of these products $pq$, where $p,q$ are the indicated distinct <primes>, its positive divisors are $1,p,q,pq$. This divisor assertion does not require unique factorization: if $d\mid pq$ and $p\nmid d$, then $p\mid pq/d$ by the <Euclid lemma> proved in (a), so $d\mid q$ and $d$ is either one or $q$. If $p\mid d$, write $d=pe$ and obtain $e\mid q$, giving $d=p$ or $pq$.
None of $3,7,13,17$ is congruent to one modulo ten. Consequently no listed product has a proper nonunit divisor in $S$. Thus each is an <irreducible element of a commutative monoid>, and the displayed equality proves the claim. Ordinary prime decompositions and irreducibility relative to the smaller <commutative monoid> $S$ are different notions.
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