Argentine National Olympiad 2015

The answer is yes for part a) and no for part b).

Consider a number $a>1$ of type 1. If $a$ is a 1-digit number then $a=(a-1)+1$ is the desired representation since $a-1$ and $1$ are 1-digit numbers, hence type 2 numbers by definition.

Let $a$ have at least two digits. Write it in the form $a=u_1{v}_1{u}_2{v}_2{u}_3{v}_3\dots$ where $u_1,u_2,\dots$ and $v_1,v_2,\dots$ are its digits at odd and even position respectively. We have $u_1>0$ for the first digit $u_1$, also $v_1\ge u_1>0$ since $a$ is of type 1.

Now let $b=u_{1}0u_{2}0u_{3}0\dots$ be the number obtained by replacing all digits at even positions by $0$. This is a type 2 number because $u_i\ge 0$ for all $i$.

Next, construct number $c$ as follows: delete the first digit $u_1$ from $a$, then replace all remaining digits $u_2,u_3,\dots$ at odd positions by zeros. In other words $c=v_{1}0v_{2}0v_{3}0\dots$; the first digit $v_1$ is nonzero. Clearly $c$ is also of type 2, and it has one digit less than $b$ (and a). Now it follows from the rule of addition that $a=b+c$, so part a) is done.

For part b) we show that the type 2 number $109$ is not representable as the sum of two type 1 numbers. Suppose on the contrary that such a representation exists. If one summand is among $1,\dots ,9$ then the other is among $100,\dots ,108$, and the latter numbers are not of type 1. So let both summands be 2-digit numbers, and let $10u+v$ be one of them, with $0<u\le v\le 9$ as $10u+v$ is of type 1. The other summand is then $b=109-(10u+v)=10(10-u)+(9-v)$. Because $1\le 10-u\le 9$ and $0\le 9-v\le 8$, the digits of $b$ are $10-u$ and $9-v$ in this order. However $10-u>9-v$ because $u\le v$, hence $b$ is not a type 1 number, contrary to the assumption.