Problem 822

Solution 1:

We are being asked to find the number of multiples of $18$ between $111\,000$ and $111\,999$ inclusive.

Since $18\times 6166=110\,988$ and
$18\times6167=111\,006$, then the
smallest multiple of $18$ greater
than or equal to $111\,000$ is $111\,006$. (We can find the numbers $6166$ and $6167$ by dividing $111\,000$ by $18$ and rounding the result both down and
up to the nearest integer.)

Since $18\times 6222=111\,996$ and
$18\times6223=112\,014$, then the
largest multiple of $18$ less than or
equal to $111\,999$ is $111\,996$. (We can similarly find the
numbers $6222$ and $6223$ by dividing $111\,999$ by $18$ and rounding the result both down and
up to the nearest integer.)

All multiples of $18$ from $18\times6167$ to $18\times6222$ inclusive satisfy the given
conditions, and so there are $6222-6167+1=56$ different possibilities
for $N$.

Solution 2:

Since $N=111\,abc$ is divisible by
$18$, then $N$ is divisible by both $2$ and $9$ since $2$ and $9$ have no factors in common and $2\times9=18$.

Since $N$ is divisible by $2$, then $N$ is even and so its units digit, $c$, must equal $0$, $2$, $4$, $6$, or $8$.

An integer is divisible by $9$
exactly when the sum of its digits is divisible by $9$.

Thus, the sum of the digits of $N$,
which is equal to $1+1+1+a+b+c=a+b+c+3$, must be a multiple
of $9$.

The smallest possible value of $a+b+c+3$ is $3$ (when $a=b=c=0$) and its largest possible value
is $30$ (when $a=b=c=9$).

The multiples of $9$ in this range
are $9$, $18$ and $27$, and so $a+b+c=6$ or $a+b+c=15$ or $a+b+c=24$.

To summarize to this point, $N=111\,abc$ is divisible by $18$ exactly when $c$ is equal to $0$, $2$, $4$, $6$, or $8$, and $a+b+c=6$ or $a+b+c=15$ or $a+b+c=24$.

We proceed by setting $c$ equal to
each of the possible even digits and determining the number of possible
values for $a$ and $b$.

When $c=0$, we get $a+b=6$ or $a+b=15$ or $a+b=24$.

When $a+b=6$, the possible values for
the digits $a$ and $b$ (written as ordered pairs $(a,b)$) are $(0,6)$, $(1,5)$, $(2,4)$, $(3,3)$, $(4,2)$, $(5,1)$, and $(6,0)$. (Alternately, when $a+b=6$, $a$ can equal any integer from $0$ to $6$ inclusive and then $b=6-a$.)

Thus, there are $7$ ordered pairs of
digits $(a,b)$ when $c=0$ and $a+b=6$, and so there are $7$ possible values of $N$.

When $a+b=15$, the possible pairs of
digits $(a,b)$ are $(6,9)$, $(7,8)$, $(8,7)$, and $(9,6)$, and so there are $4$ possible values of $N$.

When $a+b=24$, there are no possible
pairs of digits $(a,b)$ since $a+b$ is at most $9+9=18$.

When $c=2$, we get $a+b=6-2=4$ or $a+b=15-2=13$ or $a+b=24-2=22$.

We continue to count the number of pairs of digits $(a,b)$ given each of the possible values
of $c$ and summarize those results in
the tables below.

Therefore, the number of possibilities for $N$ is $11+11+11+12+11=56$.