5. A positive integer $n$ is said to be increasing if, by reversing the digits of $n$, we get an integer larger than $n$. For example, 2003 is increasing because, by reversing the digits of 2003, we get 3002, which is larger than 2003. How many four-digit positive integers are increasing?
(1 mark)
5. Let $n$ be a positive integer. When we reverse the digits of $n$, we may get an integer larger than $n$. Such an $n$ is called an "increasing number". For example, when we reverse the digits of 2003, we get 3002; and 3002 is larger than 2003, so 2003 is an "increasing number". How many four-digit positive integers are increasing?
(1 mark)
5. 4005
5. First we count the number of 'increasing' integers of the form $\mathrm{ABCA}$. In this case $\mathrm{C}$ must be greater than $\mathrm{B}$. For each choice of $\mathrm{A}$ (there are 9 choices, namely, $1,2, \ldots, 9$ ), there are $C_{2}^{10}=45$ choices for $\mathrm{B}$ and $\mathrm{C}$ (as we require $\mathrm{C}$ to be greater than $\mathrm{B}$ and we can choose among 0 to 9 ). So there are $9 \times 45=405$ 'increasing' numbers of this form.
Next we count the number of 'increasing' integers of the form $\mathrm{ABCD}$, where $\mathrm{A}$ is different from $\mathrm{D}$. In this case we need only require $\mathrm{D}$ to be greater than $\mathrm{A}$, whereas $\mathrm{B}$ and $\mathrm{C}$ can be arbitrary. Clearly, $\mathrm{A}$ and $\mathrm{D}$ have to be non-zero. So there are $C_{2}^{9}=36$ ways to choose $\mathrm{A}$ and $\mathrm{D}$, and $10^{2}=100$ ways to choose $\mathrm{B}$ and $\mathrm{C}$. Thus there are $36 \times 100=3600$ 'increasing' numbers of this form.
It follows that the answer is $405+3600=4005$.
Alternative Solution
Clearly, an increasing number cannot end with 0. Consider all four-digit positive integers that do not end with 0. There are $9 \times 10 \times 10 \times 9=8100$ such numbers. Among these, $9 \times 10=90$ of them are palindromes (as all palindromes are of the form ABBA where $\mathrm{A}$ is from 1 to 9 while $\mathrm{B}$ is from 0 to 9). So $8100-90=8010$ of them are not palindromes. By grouping an integer with its reverse (i.e. ABCD with DCBA), we get $8010 \div 2=4005$ pairs. In each pair, exactly one integer is increasing. So the answer is 4005.