For integers $a, b, c$ and $d$, it holds that $a>b>c>d$ and

$$
(1-a)(1-b)(1-c)(1-d)=10
$$

What values can the expression $a+b-c-d$ take?
I/1. If the numbers $a, b, c$ and $d$ are distinct, the numbers $1-a, 1-b, 1-c$ and $1-d$ are also distinct. Since 10 is the product of only two prime numbers, we can write 10 as the product of four integers only if two of these integers are 1 and -1. The remaining numbers are then -2 and 5 or 2 and -5.

Given $a>b>c>d$, it follows that $1-a<1-b<1-c<1-d$. Therefore, $1-b=-1$ and $1-c=1$, which gives $b=2$ and $c=0$. If $1-a=-2$ and $1-d=5$, we get $a=3$ and $d=-4$. In the case $1-a=-5$ and $1-d=2$, we have $a=6$ and $d=-1$.

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