What are the last two digits of the number

$$
8+88+888+\cdots+\overbrace{88 \cdots 88}^{2008} ?
$$
Solution 1: Since we only want to know the last two digits, it is enough to know the last two columns of this sum (the tens and the units), that is,

$$
8+88 \times 2007=8+\ldots 16
$$

Since $8+16=24$, the last two digits of the number are 24.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-097.jpg?height=389&width=699&top_left_y=1182&top_left_x=1181)

Solution 2: Observe that the last two digits of the given number are the same as the last two digits of the number

$$
8+\overbrace{88+\cdots+88}^{2007}=8+2007 \times 88
$$

which also coincide with the last two digits of $8+7 \times 88=624$. Therefore, the last two digits of the number we are looking for are 24.