Write $\alpha=a/q+\beta$, where $|\beta|\leq q^{-2}$. Split the $m$-interval into $O(M/q+1)$ consecutive blocks of length at most $q/2$. If distinct integers $m,n$ lie in one block, then $0<|m-n|<q/2$. Since $(a,q)=1$,
$$
\left\|\frac{a(m-n)}q\right\|\geq\frac1q,
$$
whereas $|\beta(m-n)|\leq1/(2q)$. Hence $\|\alpha m-\alpha n\|\geq1/(2q)$.

The points $\alpha m$ in one block are therefore $1/(2q)$-separated modulo one. Order them by distance from the nearest integer. Apart from a bounded number of endpoints, the $j$th closest point has distance $\gg j/q$, and consequently
$$
\sum_{m\text{ in one block}}\min(R,\|\alpha m\|^{-1})
\ll R+q\sum_{1\leq j\leq q}\frac1j
\ll R+q\log q.
$$
Multiplying by $O(M/q+1)$ gives
$$
\boxed{
\sum_{m_0\leq m<m_0+M}\min(R,\|\alpha m\|^{-1})
\ll(\log q)\left(\frac{MR}{q}+M+R+q\right).}
$$
This is the <reciprocal fractional-part sum near a rational> estimate.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 117 / 3 / b / Solution

Write $M_0=CRT^2$ and split the last sum in part d according as $\tau_4(m)<X$ or $\tau_4(m)\geq X$. For the first part, the <reciprocal fractional-part sum near a rational> gives
$$
\sum_{\substack{m\leq M_0\\\tau_4(m)<X}}
\tau_4(m)\min(R,\|\alpha m\|^{-1})
\ll X(\log q)\left(\frac{M_0R}{q}+M_0+R+q\right).
$$
For the second part, use the supplied second-moment estimate and <truncation of a divisor weight by its second moment>:
$$
\begin{aligned}
\sum_{\substack{m\leq M_0\\\tau_4(m)\geq X}}
\tau_4(m)\min(R,\|\alpha m\|^{-1})
&\leq R\sum_{\substack{m\leq M_0\\\tau_4(m)\geq X}}\tau_4(m)\\
&\ll\frac{RM_0}{X}(\log M_0)^{O(1)}.
\end{aligned}
$$
Substitute $M_0\asymp RT^2$, take fourth roots, and factor out $R^{1/2}T^{1/2}$. The four terms from the bounded-weight estimate become, after harmless enlargement by $(\log q)(\log RT)^{O(1)}$,
$$
\frac{X^{1/4}}{q^{1/4}},qquad
\frac{X^{1/4}}{R^{1/4}},qquad
\frac{X^{1/4}q^{1/4}}{R^{1/2}T^{1/2}},
$$
with the smaller $XR$ term absorbed by $XM_0$. The large-weight part contributes $X^{-1/4}$. Finally, the two diagonal terms from part d contribute $R^{-1/4}$ and $T^{-1/4}$; since $X\geq1$, the first is absorbed by $X^{1/4}R^{-1/4}$. Therefore
$$
\boxed{
|S|\ll\|b\|_2\|c\|_2(\log q)(\log RT)^{O(1)}R^{1/2}T^{1/2}
\left(
\frac1{X^{1/4}}+\frac{X^{1/4}}{q^{1/4}}
+\frac{X^{1/4}}{R^{1/4}}
+\frac{X^{1/4}q^{1/4}}{R^{1/2}T^{1/2}}
+\frac1{T^{1/4}}
\right).}
$$

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 117 / 3 / e / Solution

If $(a,q)=1$, $|\alpha-a/q|\leq q^{-2}$, and $M,R\geq2$, then uniformly in $m_0$,
$$
\sum_{m_0\leq m<m_0+M}\min(R,\|\alpha m\|^{-1})
\ll(\log q)\left(\frac{MR}{q}+M+R+q\right).
$$
Split the interval into blocks of length at most $q/2$. Within a block the points $\alpha m$ are $\gg1/q$-separated modulo one; ordering their distances from the nearest integer gives $O(R+q\log q)$ per block.

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