Problem 523

Every multiple of $21$ is of
the form $21k$ for some integer $k$.

For such a multiple to be between 10 000 and 100 000, we need $10\,000 < 21k < 100\,000$ or $\$\frac{10\,000}{21}$ < k <
$\frac{100\,000}{21}\$.$

Since $\$\frac{10\,000}{21} \approx$
476.2$and$$\frac{100\,000}{21} \approx$ 4761.9$and$k$ is an
integer, then $477 \leq k \leq 4761$.
(Note that $k$ is greater than $476.2$ and is an integer, so must be at
least $477$; similarly, $k$ is at most $4761$.)

We also want the units digit of $21k$
to be $1$.

This means that the units digit of $k$ itself is $1$, since the units digit of the product
of $21$ and $k$ is equal to units digit of $k$ because the units digit of $21$ is $1$.

Therefore, the possible values of $k$
are $481, 491, 501, $\ldots,$ 4751,
4761$.

There are $429$ such values. To see
this, we can see that counting the integers in this list is the same as
counting the integers in the list $48, 49,
50, $\ldots,$ 475, 476$. This list is equivalent to removing the
integers from $1$ to $47$ from the list of integers from $1$ to $476$, giving $476 - 47 = 429$ integers.

Thus, $M = 429$.
Solution 1:

We can partition the $N$ students
at Strickland S.S. into four groups:

$a$ students who are in the
physics club and are in the math club
$b$ students who are in the
physics club and are not in the math club
$c$ students who are not in
the physics club but are in the math club
$d$ students who are not in
the physics club and are not in the math club

In Math Club
Not in Math Club

In Physics Club
$a$
$b$

Not in Physics Club
$c$
$d$

From the given information, there are $\frac{2}{5}N$ students in the physics
club. In other words, $a+b =
$\frac{2}{5}N\$.$

Among the students in the physics club, twice as many are not in the
math club as are in the math club. This means that $b = $\frac{2}{3} \cdot \frac{2}{5}N = \frac{4}{15}N$$and$a =
$\frac{1}{3}\cdot \frac{2}{5}N = \frac{2}{15}N\$.$

From the given information, there are $\frac{1}{4}N$ students in the math club.
In other words, $a + c =
$\frac{1}{4}N\$.$

Since $a = \frac{2}{15}N$, then $c = $\frac{1}{4}N - \frac{2}{15}N = \frac{7}{60}N\$.$

Since $a + b + c + d = N$, then $d = N - a - b - c = N - $\frac{4}{15}N - \frac{2}{15}N - \frac{7}{60}N = \frac{29}{60}N\$.$

Lastly, we know that $500 < N <
600$.

Since each of $a$, $b$, $c$, and $d$ is an integer, then $N$ must be divisible by 60.

Therefore, $N = 540$ and so the
number of students not in either club is $d =
$\frac{29}{60} \cdot$ 540 = 261$.

Solution 2:

Since there are $N$ students at
Strickland S.S., then $\frac{2}{5}N$
are in the physics club and $\frac{1}{4}N$ are in the math club.

Since each of $\frac{2}{5}N$ and
$\frac{1}{4}N$ must be an integer,
then $N$ must be divisible by $5$ and must be divisible by $4$.

Since $5$ and $4$ share no common divisor larger than
$1$, then $N$ must be divisible by $5 \cdot 4 = 20$.

Thus, we let $N = 20m$ for some
positive integer $m$.

In this case, $\frac{2}{5}N = 8m$
students are in the physics club and $\frac{1}{4}N = 5m$ students are in the
math club.

Now, among the $8m$ students in the
physics club, twice as many are not in the math club as are in the math
club.

In other words, $\frac{1}{3}$ of the
$8m$ students in the physics club are
in the math club. This means that $m$
must be divisible by $3$, since $3$ is a prime number and $8$ is not divisible by $3$.

Therefore, $m = 3k$ for some positive
integer $k$, which means that $N = 20m = 60k$ and $\frac{2}{5}N = 8m = 24k$ and $\frac{1}{4}N = 5m = 15k$.

Since $500 < N < 600$ and $N$ is a multiple of 60, then $N = 540$, which means that $k = 9$.

Thus, the number of students in the physics club is $24k = 216$, of whom $\frac{1}{3} \cdot 216 = 72$ are in the
math club and $\$\frac{2}{3} \cdot$ 216=
144$ are not in the math club.

Also, the number of students in the math club is $15k = 135$.

Finally, we know that

there are $540$ students at
the school,
$72$ of whom are in both the
physics club and the math club,
$144$ of whom are in the
physics club and not in the math club, and
$135 - 72 = 63$ are in the
math club and not in the physics club.

Therefore, the number of students in neither club is $540 - 72 - 144 - 63 = 261$.