Another proof for how many n-digit numbers are even For n is greater than or equal to 2, how many n-digit numbers are even. Give at least 2 proofs.
I did this proof:
The first digit cannot be zero (9 choices) and the last digit has to be even (5 choices). This leaves n-2 places in the number, each of which has 10 choices. Therefore, $9*10^{(n-2)}*5$ is the required number.
Can someone help me work a second proof? I considered complementary counting, but that would be the exact same thing basically and probably is not viable for a second proof.
 A: Well, simplest and obvious proof would be to note there are $10^n - 10^{n-1} = 9*10^{n-1}$ n-digit numbers.  Half of them are even so there are $45*10^{n-2}$ even numbers.
Well, here's another argument.  There are $10^n$ non-negative integers (including zero) that are less than $10^n$.  Half of them, $5*10^n$, are even.  Therefore there are $5*10^{n-1}$ even numbers less than $10^{n-1}$.  So there are $5*10^n - 5*10^{n-1} = 45*10^{n-2}$ even numbers at least $10^{n-1}$ and less than $10^n$.
But I'm not sure these proofs are really any different.
A: The easiest way to show that half the numbers are even is to pair them up.
If $x$ is even, then consider the ordered pair $(x, x+1)$. Every even number appears as the first value of the ordered pair it defines, as for even $x$, $10^{n-1}\leq x<10^n\Rightarrow 10^{n-1}\leq x+1<10^n$, so the second number of the pair is always an odd number which has $n$ digits.
If $x$ is odd and has $n$ digits, it is easy to see $x-1$ is even and also has $n$ digits. Clearly it follows that $x$ is in the pair $(x-1, x)$.
Since all numbers are paired up with one even and one odd number in each pair, it can be seen that half of the $n$-digit numbers are even.
Another way to pair is to pair numbers with a constant sum.
Consider pairs of numbers summing to $10^n+10^{n-1}-1$, with the first number less than the second number. One such pair is: $(10\dots0, 99\dots9)$. As the sum of the 2 numbers is odd, it can be shown that one number has to be even, and the other number has to be odd.
For a number $x$, it is either in the pair $(x, 10^n+10^{n-1}-1-x)$ or $(10^n+10^{n-1}-1-x, x)$, depending on which of the 2 numbers in the pair is bigger. Hence, every number gets paired up with another number of a different parity. As such, it follows that by considering all the pairs, half of the numbers are even.
After proving that half the numbers are even, it suffices to complete the calculation with $\frac{10^n-10^{n-1}}{2}$, which is half the total number of numbers with $n$ digits.
