9. On the coordinate plane a point whose $x$ - and $y$-coordinates are both integers is said to be a 'lattice point'. If a lattice point is randomly chosen inside (excluding the boundary) the square with vertices $(0,0),(101,0),(101,101)$ and $(0,101)$, what is the probability that the line segment (excluding the two endpoints) joining $(0,2010)$ and the point chosen contains an even number of lattice points?
(1 mark
在座標平面上， $x$ 、 $y$ 座標皆是整數的點稱爲「格點」。現於頂點爲 $(0,0)$ 、 $(101,0)$ 、 $(101,101)$ 和 $(0,101)$ 的正方形內（不包括邊界）隨意選一格點, 那麼連起 $(0,2010)$ 和所選格點的線段上（不包括兩個端點）有偶數個格點的概率是多少?
(1 分)
9. $\frac{3}{4}$
9. Note that on the line segment joining $(a, b)$ and $(c, d)$, where $a, b, c, d$ are integers, the number of lattice points (excluding the endpoints) is exactly 1 less than the H.C.F. of $c-a$ and $d-b$.
Hence there is an even number of lattice points on the line segment (excluding the endpoints) joining $(0,2010)$ and $(a, b)$ if and only if the H.C.F. of $a$ and $2010-b$ is odd, which is true if and only if at least one of $a, b$ is odd.
Now there are 100 choices for each of $a$ and $b$ (namely, $1,2, \ldots, 100$ ). Hence the probability that both are even is $\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$, and so the required probability is $1-\frac{1}{4}=\frac{3}{4}$.