, 2026

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 $(x,y) = (2,10)$.