Exercise 2.2.6 (Number of solutions of some congruences)

Question

How many solutions are there to each of the following congruences
\begin{enumerate}
\item[(a)] $15x \equiv 25 \pmod{35}$;
\item[(b)] $15 x \equiv 24 \pmod {35}$;
\item[(c)] $15 x \equiv 0 \pmod{35}$;
\end{enumerate}

Richard Ganaye · Accepted Answer

\begin{proof}
We count the numbers of solutions modulo $35$.
\begin{enumerate}
\item[(a)] $15x \equiv 25 \pmod{35} \iff 3 x \equiv 5 \pmod 7 \iff x \equiv 4 \pmod 7$.
The solutions modulo $35$ are $4, 11, 18, 25, 32$ : $5$ solutions.

\item[(b)] $15 x \equiv 24 \pmod 35$

Since $\mathrm{gcd}(15,35) = 5 
mid 24$, there is no solution: $0$ solution.

\item[(c)] $15 x \equiv 0 \pmod{35} \iff 3 x \equiv 0 \pmod 7 \iff x \equiv 0 \pmod 7$.
The solutions modulo $35$ are $0,7,14,21,28$: $5$ solutions.
\end{enumerate}
\end{proof}

Exercise 2.2.6 (Number of solutions of some congruences)

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Exercise 2.2.6 (Number of solutions of some congruences)

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