Problem 18

We look for integers between $2026$ and $2100$ that have a sum of digits of $10$.

Such integers are of the form $20xy$
for some digits $x$ and $y$.

Since the sum of the digits is $10$,
then $2 + 0 + x + y = 10$ or $x + y = 8$.

For $20xy$ to be greater than $2026$, we need $x \geq 2$.

If $x = 2$, then $y = 6$, which gives us the integer
$2026$.

To find the next greatest integer, we try $x
= 3$, which gives$y = 5$.
This gives us the integer $2035$,
which is the smallest integer $n >
2026$ with the desired property.
Consider the three-digit positive integers $abc$ with product of digits equal to $9$ (that is, $a
$\cdot b \cdot$ c = 9$).

Since $9 = 3^2$, then $a$, $b$
and $c$ are either $1$, $1$, $9$
in some order or $1$, $3$, $3$
in some order. (There are no other factors that can be used.)

The possible integers are $119$,
$191$, $911$, $133$, $313$, $331$; there are $6$ such integers.
Since the sum of $x$, $3x$ and $4y$ is equal to 48, then $x + 3x + 4y = 48$ and so $4x + 4y = 48$ or $x + y = 12$.

Since the average of $x$ and $y$ is equal to $3x$, then $\dfrac{x+y}{2} = 3x$ and so $x+y = 6x$ or $y
= 5x$.

Substituting $y = 5x$ into $x + y = 12$, we obtain $x + 5x = 12$ or $6x = 12$ and so $x = 2$.

Since $y = 5x$, then $y = 10$ and so $(x,y) = (2,10)$.