CEMC Fermat, 2023

We write an integer $n$ with
$100 \leq n \leq 999$ as $n = 100a + 10b + c$ for some digits $a$, $b$
and $c$.

That is, $n$ has hundreds digit $a$, tens digit $b$, and ones digit $c$.

For each such integer $n$, we have
$s(n) = a + b + c$.

We want to count the number of such integers $n$ with $7 \leq a+b+c \leq 11$.

When $100 \leq n \leq 999$, we know
that $1 \leq a \leq 9$ and $0 \leq b \leq 9$ and $0 \leq c \leq 9$.

First, we count the number of $n$
with $a+b+c=7$.

If $a=1$, then $b+c=6$ and there are 7 possible pairs of
values for $b$ and $c$. These pairs are $(b,c) = (0,6),(1,5),(2,4),(3,3),(4,2),(5,1),(6,0)$.

If $a=2$, then $b+c=5$ and there are 6 possible pairs of
values for $b$ and $c$.

Similarly, when $a=3, 4, 5, 6, 7$,
there are 5, 4, 3, 2, 1 pairs of values, respectively, for $b$ and $c$.

In other words, the number of integers $n$ with $a+b+c=7$ is equal to $7+6+5+4+3+2+1=28$.

Using a similar process, we can determine that the number of such
integers $n$ with $s(n) = 8$ is $8+7+6+5+4+3+2+1=36$ and the number of such
integers $n$ with $s(n) = 9$ is $9+8+7+6+5+4+3+2+1=45$.

We have to be more careful counting the number of integers $n$ with $s(n) = 10$ and $s(n) = 11$, because
none of the digits can be greater than 9.

Consider the integers $n$ with $a+b+c=10$.

If $a=1$, then $b+c=9$ and there are 10 possible pairs of
values for $b$ and $c$. These pairs are $(b,c)=(0,9),(1,8),\ldots,(8,1),(9,0)$.

If $a=2$, then $b+c=8$ and there are 9 possible pairs of
values for $b$ and $c$.

As $a$ increases from 1 to 9, we find
that there are $10+9+8+7+6+5+4+3+2=54$ such integers $n$.

(Note that when $a=9$, we have $b+c=1$ and there are 2 pairs of values for
$b$ and $c$.)

Finally, we consider the integers $n$
with $a+b+c=11$.

If $a=1$, then $b+c=10$ and there are 9 possible pairs of
values for $b$ and $c$. These pairs are $(b,c)=(1,9),(2,8),\ldots,(8,2),(9,1)$.

If $a=2$, then $b+c=9$ and there are 10 possible pairs of
values for $b$ and $c$.

If $a=3$, then $b+c=8$ and there are 9 possible pairs of
values for $b$ and $c$.

Continuing in this way, we find that there are $9+10+9+8+7+6+5+4+3 = 61$ such integers
$n$.

Having considered all cases, we see that the number of such integers
$n$ is $$S = 28 + 36 + 45 + 54 + 61 = 224$$ The rightmost two digits of
$S$ are 24.