Summation problem related to fractional part function

Question

Let $p$ and $q$ be distinct primes such that $\displaystyle \sum_{k=1}^{q} \left\{\frac{kp}{q} \right\}=51$. Find the unit digit of $q$.

Note that $\{a\}$ signifies the fractional part of $a$.

Answer 1

Let,

$S = \displaystyle \sum_{k=1}^{q} \left\{\frac{kp}{q} \right\}$

Since the summand is $0$ for $k = 0$ and $k = q$, we get,

$ S = \displaystyle \sum_{k=0}^{q} \left\{\frac{kp}{q} \right\} $

Change $k$ to $q-k$

$ \displaystyle \implies S = \sum_{k=0}^{q} \left\{p-\frac{kp}{q} \right\} $

$ \displaystyle = \sum_{k=1}^{q-1} \left\{-\frac{kp}{q} \right\} $

Note that,

$ \{x\} + \{-x\} = 1 \ ; \ \forall \ x \notin \mathbb{Z} $

$ \implies S = \displaystyle \sum_{k=1}^{q-1} \left(1 - \left\{\frac{kp}{q} \right\}\right) $

$ \implies S = q-1 - S $

$ \implies S = \dfrac{q-1}{2} $

Since $S = 51$

$ \implies q = 103 $

$ \implies q \pmod {10} = 3 $