Slovenia — National Math Olympiad

The numbers $a, b, c$ and $d$ are pairwise different, so $1-a, 1-b, 1-c$ and $1-d$ are all different as well. Since $10$ is the product of two primes, it can only be written as the product of four integers if two of these integers are $1$ and $-1$. The remaining two factors are either $-2$ and $5$ or $2$ and $-5$.

Since $a > b > c > d$ we have $1-a < 1-b < 1-c < 1-d$. So, $1-b = -1$ and $1-c = 1$, which implies $b=2$ and $c=0$. If $1-a = -2$ and $1-d = 5$, we get $a=3$ and $d=-4$. In the case where $1-a = -5$ and $1-d = 2$, we have $a=6$ and $d=-1$.

The value of $a+b-c-d$ is equal to $3+2-0-(-4) = 9$ in the first case, and to $6+2-0-(-1) = 9$ in the second case, so there is really only one possibility, $a+b-c-d=9$.