, 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 $a = u_1 v_1 u_2 v_2 u_3 v_3 \ldots$ where $u_1, u_2, \ldots$ and $v_1, v_2, \ldots$ are its digits at odd and even position respectively. We have $u_1 > 0$ for the first digit $u_1$, also $v_1 \geq u_1 > 0$ since $a$ is of type 1.

Now let $b = u_1 0 u_2 0 u_3 0 \ldots$ be the number obtained by replacing all digits at even positions by $0$. This is a type 2 number because $u_i \geq 0$ for all $i$.

Next, construct number $c$ as follows: delete the first digit $u_1$ from $a$, then replace all remaining digits $u_2, u_3, \ldots$ at odd positions by zeros. In other words $c = v_1 0 v_2 0 v_3 0 \ldots$; 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 $1, \ldots, 9$ then the other is among $100, \ldots, 108$, 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 $0 < u \leq v \leq 9$ as $10u + v$ is of type 1. The other summand is then $b = 109 - (10u + v) = 10(10 - u) + (9 - v)$. Because $1 \leq 10 - u \leq 9$ and $0 \leq 9 - v \leq 8$, the digits of $b$ are $10 - u$ and $9 - v$ in this order. However $10 - u > 9 - v$ because $u \leq v$, hence $b$ is not a type 1 number, contrary to the assumption.