= Solution
For each <prime number> $p$, reduction of the <mapping cone> is the <mapping cone> of the reduced <chain map>. Part (a), whose proof works over any coefficient ring, therefore gives
$$
H_i(M\otimes\mathbb F_p)=0\qquad\text{for every }i,p.
$$
Every $M_i$ is a finitely generated <free abelian group>. Multiplication by $p$ consequently gives a <short exact sequence of chain complexes>
$$
0\longrightarrow M\xrightarrow{p}M\longrightarrow M\otimes\mathbb F_p\longrightarrow0.
$$
The neighboring terms $H_{i+1}(M\otimes\mathbb F_p)$ and $H_i(M\otimes\mathbb F_p)$ in its <long exact sequence in homology> both vanish. Thus multiplication by $p$ on $H_i(M)$ is an <isomorphism> for every <prime number> $p$.
The group $H_i(M)$ is a <finitely generated abelian group>, since it is a quotient of a subgroup of $M_i$. Write it, using the <Fundamental theorem of finitely generated abelian groups>, as
$$
H_i(M)\cong\mathbb Z^{b_i}\oplus T_i,
$$
with $T_i$ finite. If $b_i>0$, multiplication by any <prime number> is not surjective. If $T_i\ne0$, choose a <prime number> dividing its order; multiplication by that prime is not injective. Both contradict the preceding <isomorphism>. Hence $H_i(M)=0$ for every $i$.
Applying part (a) integrally now proves the stronger natural conclusion
$$
\boxed{f_*:H_i(C;\mathbb Z)\xrightarrow{\ \cong\ }H_i(D;\mathbb Z)\quad\text{for every }i.}
$$
This is <prime coefficient detection of quasi-isomorphisms>. The finite-generation hypothesis is used precisely when excluding nonzero groups on which every prime acts invertibly.
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