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Use $N,p,a,M,S,D$ and the full profile from L13, with $D>0$. Each nonunit divisor of $N$ has a unique form $p^jm$, where $0\le j\le a$, $m\mid M$, and $(j,m)\ne(0,1)$. The cylinders are

$$Y_t=\{x\in\mathbb Z/N\mathbb Z:x\equiv t\pmod{p^a}\}, \qquad t\in\mathbb Z/p^a\mathbb Z.$$

They partition $\mathbb Z/N\mathbb Z$ into $p^a$ sets of $M$ points. All masses below count class incidences, with overlaps counted repeatedly.

The exact contribution of one class

Since $\gcd(p^a,m)=1$ for every $m\mid M$, prescribed residues modulo $p^a$ and $m$ select exactly one class modulo $p^am$. To see this, write $x=t+p^az$: multiplication by $p^a$ is invertible modulo $m$, so the congruence modulo $m$ chooses exactly one residue of $z$ modulo $m$. There are $N/(p^am)=M/m$ such points modulo $N$. This includes $m=1$, when the congruence modulo $m$ imposes no restriction.

For a class with modulus $m>1$ coprime to $p$, the count is $M/m$ in every cylinder. The sum of these contributions is

$$\sum_{\substack{m\mid M\\m>1}}\frac{M}{m} =\sum_{d\mid M}d-M=S-M.$$

Here $m\mapsto M/m$ permutes the divisors of $M$, and the omitted modulus $m=1$ accounts for the subtracted term $M$.

Next take a class $b_{p^jm}(p^jm)$ with $1\le j\le a$ and $m\mid M$. It meets $Y_t$ precisely when

$$t\equiv b_{p^jm}\pmod{p^j}.$$

There are exactly $p^{a-j}$ such residues $t$ modulo $p^a$. On each of these cylinders the congruence modulo $p^j$ is automatic, and the remaining condition modulo $m$ selects exactly $M/m$ points by the coprime count. Thus the contribution has weight $M/m$ on exactly $p^{a-j}$ cylinders, including just one cylinder when $j=a$.

Necessary loads and total supply

For each $t$, let $T_t$ be the total contribution in $Y_t$ from the classes whose moduli are divisible by $p$. The full profile's incidence mass there is

$$(S-M)+T_t.$$

Consequently, even the necessary cylinder mass inequalities for a cover require

$$(S-M)+T_t\ge M,\qquad\hbox{that is,}\qquad T_t\ge D \quad\hbox{for every }t.$$

Summing the exact class-support counts gives a total independent of all residue choices:

$$\sum_t T_t =\sum_{j=1}^a\sum_{m\mid M}\frac{M}{m}p^{a-j} =S\sum_{j=1}^a p^{a-j} =S\frac{p^a-1}{p-1}.$$

Subtracting forced excess

Call a contribution heavy if its weight $M/m$ exceeds $D$. Define threshold waste and useful supply by

$$W_t=(T_t-D)_+=\max(T_t-D,0),\qquad V=\sum_t\min(T_t,D)=\sum_t T_t-\sum_t W_t.$$

These are quantities for the cylinder lower-bound constraints. They do not count distinct covered points or identify the actual holes left by the classes coprime to $p$.

At a fixed cylinder, suppose there are $h\ge1$ heavy contributions of weights $w_1,\ldots,w_h>D$. All other contributions are nonnegative. At least one heavy contribution is present, each exceeds $D$, and all weights are positive. Thus $T_t\ge\sum_{\ell=1}^h w_\ell>D$, and hence

$$W_t=T_t-D \ge\sum_{\ell=1}^h w_\ell-D =\sum_{\ell=1}^h(w_\ell-D)+(h-1)D \ge\sum_{\ell=1}^h(w_\ell-D).$$

The last inequality uses $(h-1)D\ge0$, since $D>0$.

If $h=0$, the last sum is empty and equals zero, and $W_t\ge0$ gives the same lower bound. Thus any overlap between heavy contributions preserves the bound. Contributions of weight at most $D$ cannot decrease $W_t$.

Each heavy class of modulus $p^jm$ contributes its forced excess $M/m-D$ on exactly $p^{a-j}$ cylinders. Summing the last inequality over all cylinders yields

$$\sum_t W_t\ge \sum_{\substack{1\le j\le a,\ m\mid M\\M/m>D}} (M/m-D)p^{a-j}.$$

Combining this with the total supply gives $V\le U$. The equivalent formula for $U$ in the statement follows by applying $w-(w-D)_+=\min(w,D)$ separately to every weight $w=M/m$ in the finite sum.

If every cylinder meets its necessary load $T_t\ge D$, then $\min(T_t,D)=D$ in all $p^a$ cylinders, so

$$Dp^a=V\le U.$$

This contradicts $U<Dp^a$. Therefore that strict inequality rules out every full-profile placement satisfying even the cylinder mass constraints, and in particular rules out every full-profile cover.

Distinct subprofiles and the guards

Suppose a family with pairwise distinct nonunit moduli dividing $N$ were a cover. Add one arbitrarily placed residue class for each unused nonunit divisor of $N$. The resulting full profile still covers, and no cylinder incidence mass has decreased. This contradicts the full-profile exclusion. The same argument shows directly that feasibility of the cylinder lower bounds is upward closed under adding missing classes. A cover of higher multiplicity is also a one-cover and is excluded in this domain. Repeated moduli are not subprofiles of the full inventory, so the argument makes no assertion about them.

If $D\le0$, the constant contribution $S-M$ is already at least $M$, and these full-profile cylinder lower bounds supply no obstruction. For $D>0$, the proof only gives $Dp^a\le U$ as a necessary condition. Equality or the reverse strict inequality therefore does not decide feasibility or coverage.

For the boundary $M=1$, the divisor sum is $S=1$ and $D=1$. There are no nonunit moduli coprime to $p$, and every class of modulus $p^j$ has weight one. There are no heavy terms. Thus

$$\frac{U}{p^a}=\sum_{j=1}^a p^{-j} =\frac{1-p^{-a}}{p-1}<1,$$

because $p\ge2$. The strict criterion applies at every depth $a\ge1$, including $a=1$, and excludes distinct nonunit-divisor covers of a pure prime-power frame. No finite enumeration is used.