Estonian Mathematical Olympiad

Let the sum of the digits with a plus sign in front of them be $x$ and the absolute value of the sum of the digits with a minus sign in front of them be $y$. Then the value $v$ of the expression equals $x-y$. Also $x+y=0+1+2+3+4+5+6+7+8+9=45$. Therefore $v=45-2y$.

a) As $45-2y\le 45$, nothing greater than $45$ can be the value of the expression. We now show that we can obtain all the positive odd integers up to $45$ as the result of the expression; this shows that the least positive integer that cannot be equal to the result is $47$.

We previously showed that $y=\frac{45-v}{2}$. In order to make the value of the expression be $v$, we need to put a minus sign in front of some digits that sum up to $\frac{45-v}{2}$. As $v$ is a positive odd integer between $1$ and $45$, the number $\frac{45-v}{2}$ is a nonnegative integer between $0$ and $22$. Each such positive integer can be written as a sum of digits as follows: numbers from $1$ to $9$ are among the digits themselves, numbers from $10$ to $17$ can be obtained as the sum of $9$ and some other digit, and numbers $18$ to $22$ can be written as the sum of $9$, $8$ and some other digit in the range of $1$ to $5$. The case $y=0$ corresponds to the version where every digit has a plus sign in front of it.

b) Number $2$ cannot be obtained as the value of the expression, because solving $2=45-2y$ gives $y=21.5$, which is impossible in integers. The number $2$ is also the least positive even number.