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Source. Expansion of the cyclic specialization of the corollary on printed p. 378 (Part I PDF p. 3) and the unexpanded transfer in Part II on printed p. 79 (Part II PDF p. 4). This is a complete compilation-supplied deduction from the finite Chinese remainder theorem. Part I only needs product sets with the correct projection cardinalities; the explicit digit reversal below supplies the aligned intervals required by Part II's formulation.

Statement

Let $N=\prod_{i=1}^n p_i^{s_i}$, with $n\ge1$, distinct primes $p_i$, and positive integer exponents $s_i$. There is a bijection

$$\Theta:\mathbb Z/N\mathbb Z\longrightarrow P=\prod_i\{0,\ldots,p_i^{s_i}-1\}$$

that sends every congruence class modulo a divisor $m=\prod_i p_i^{t_i}$ of $N$ to an aligned box whose $i$-th side is an interval of $p_i^{s_i-t_i}$ consecutive integers starting at a multiple of $p_i^{s_i-t_i}$. Every such box has a unique inverse image of this form. Its cardinality is $N/m$.

Consequently this correspondence preserves covers, properness and equality or inequality of cardinalities. Distinct moduli correspond exactly to distinct box cardinalities.

Proof

The finite Chinese remainder theorem gives the bijection

$$x\pmod N\longmapsto(x\pmod{p_1^{s_1}},\ldots, x\pmod{p_n^{s_n}}).$$

In a coordinate with prime $p$ and exponent $s$, write its least nonnegative representative uniquely as $x=\sum_{j=0}^{s-1}e_jp^j$, where $0\le e_j<p$. Define

$$\rho_{p,s}(x)=\sum_{j=0}^{s-1}e_jp^{s-1-j}.$$

This reverses the $s$ digits, including initial zero digits. Reversing twice returns $x$, so it is a bijection. Let $\Theta$ be the Chinese remainder bijection followed by this reversal in each coordinate.

The condition $x\equiv a\pmod{p^t}$ fixes precisely the low digits $e_0,\ldots,e_{t-1}$. Their reversal fixes the high $t$ digits and leaves the remaining $s-t$ digits arbitrary. If $c=\sum_{j=0}^{t-1}e_jp^{t-1-j}$, the image is exactly

$$\{cp^{s-t},\ldots,(c+1)p^{s-t}-1\}.$$

For $t=0$ take $c=0$: the image is the full coordinate. For $t=s$ it is a singleton. Applying this in all coordinates proves the forward claim and the size formula $\prod_i p_i^{s_i-t_i}=N/m$.

Conversely, an aligned interval of length $p^{s-t}$ specifies exactly those high $t$ digits; reverse them to recover a unique residue modulo $p^t$. For a product of these intervals, the Chinese remainder theorem gives one residue modulo $m=\prod_i p_i^{t_i}$. This is the unique inverse image. A finite cyclic group of order $N$ has one subgroup of each index $m\mid N$, namely $m\mathbb Z/N\mathbb Z$ after a generator is chosen, so the residue classes are precisely its cosets.

Finally, a family of integer residue classes with moduli dividing $N$ covers $\mathbb Z$ if and only if their images cover $\mathbb Z/N\mathbb Z$: every membership condition depends only on the residue modulo $N$. Bijections preserve unions, the full group corresponds to the full box, and $m\mapsto N/m$ is injective. These facts prove all the stated covering and cardinality consequences.

External input and scope

The only external theorem here is the finite Chinese remainder theorem: for pairwise coprime positive moduli, reduction from the residue ring modulo their product to the product of the residue rings is a bijection. Oddness is unnecessary for this correspondence itself. It enters the subsequent counting obstructions. Arbitrary cosets in noncyclic Sylow groups are not being identified with aligned intervals; the separate nilpotent-group proof uses Part I's broader product-set theorem.

Bears on. The geometric reductions for Problem 7. In the square-free case, all exponents are one, so these boxes become the usual CRT hyperplanes.