Exercise 1.2.5 (Number of multiples of $7$)

Question

How many integers between $100$ and $1000$ are divisible by $7$?

Richard Ganaye · Accepted Answer

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

\begin{proof} The number $N$ of integers $x$  between $100$ and $1000$ which are divisible by $7$ is the number of integers $k$ such that
\begin{align}
100 \leq 7k \leq 1000.
\end{align}
Since $7 
mid 100$, this condition is equivalent to
\begin{align}
\frac{100}{7} < k \leq \frac{1000}{7}.
\end{align}
Note that for $x \in \R$ and $k \in \Z$,
$$x < k \iff \lfloor x floor + 1 \leq k,$$
and
$$ k \leq x \iff k \leq \lfloor xfloor.$$
So the condition (2) is equivalent to
\begin{align}
\left \lfloor \frac{100}{7} ightfloor + 1 \leq k \leq \left \lfloor \frac{1000}{7} ight floor.
\end{align}
Therefore
$$N = \left \lfloor \frac{1000}{7} ight floor - \left \lfloor \frac{100}{7} ightfloor  = 142 - 14 = 128.$$
\end{proof}

\bigskip
Check:

\begin{verbatim}
 counter = 0
 for x in range(100,1001):
     if x % 7 == 0:
         counter += 1
 print(counter)
                             128
\end{verbatim}

Exercise 1.2.5 (Number of multiples of $7$)

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Exercise 1.2.5 (Number of multiples of $7$)

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