3. Find the number of pairs of consecutive integers in the set $\{1000,1001,1002, \ldots, 2000\}$ such that no carrying is required when the two integers are added.
(1 mark)
3. 在集合 $\{1000,1001,1002, \ldots, 2000\}$ 中, 有多少對連續整數加起來時不用進位?
3. 156
3. Suppose $n$ and $n+1$ are to be added, and write $n$ in the form $1 \mathrm{ABC}$. If any one of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ is equal to $5,6,7$ or 8 , then carrying is needed in the addition. If $\mathrm{B}$ is 9 and $\mathrm{C}$ is not 9 , or if $\mathrm{A}$ is 9 and $\mathrm{B}$ and $\mathrm{C}$ are not both 9 , carrying is also required. Therefore, if no carrying is required, $n$ must be of one of the following forms:
$$
1 \mathrm{ABC} \quad 1 \mathrm{AB} 9 \quad 1 \mathrm{~A} 99 \quad 1999
$$
where each of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ is from 0 to 4 . It follows that the answer is $5^{3}+5^{2}+5+1=156$.