Solution Verification: How many positive integers less than $1000$ have at least one digit that is a $9$?

Question

Here's how I solved this:

1) No restrictions: $[1, 999]$ = $999$ numbers

2) Violation (no $9$s): there are $9$ digit choices for the first, $9$ for the second, and $9$ for the third because we exclude the number $9$ itself from the range $[0, 10]$ = $729$ such numbers

3) Condition enabled: $999$ - $729$ = $270$.

Something feels off, though. Is this solution correct? Sorry if this seems trivial.

Edit: I suspect I may encounter problems for numbers where the first two digits are $0$s--did I overcount?

Answer 1

You can salvage your approach.

A positive integer less than $1000$ has a unique representation as a $3$-digit number padded with leading zeros, if needed.

To avoid a digit of $9$, you have $9$ choices for each of the $3$ digits, but you don't want all zeros, so the excluded set has count $9^3 - 1 = 728$.

Hence the count you want is $999 - 728 = 271$.

Answer 2

To consider the complement is a fine idea, but better to partition $[1,999]$ as $$ I_1 \cup I_2 \cup I_3 = [1,9]\cup[10,99]\cup[100,999] $$ in such a way that every element of $I_k$ has exactly $k$ digits. Let us call bad a number with no $9$ in its decimal representation. There are $8$ bad numbers in $I_1$, $8\cdot 9=72$ bad numbers in $I_2$ and $8\cdot 9\cdot 9$ bad numbers in $I_3$, hence a total of $8(1+9+9^2)=728$ bad numbers in $[1,999]$, contra $999-728=\color{red}{271}$ numbers with at least one $9$ in their decimal representation.

Answer 3

You can build an argument as follows. There is exactly one number less then 10. What about less than a 100? There are 19 of them: 9, 19, ..., 99 and 90, 91, ..., 98. What about less than 1000? As it is already mentioned, there are 271. In general, let $N_k$ be the number of numbers less than $10^k$ that contain the digit 9. Then $$N_{k+1} = 9N_k + 10^k = 10^k - 9^k.$$

Interestingly enough, we have $$\frac{N_{k+1}}{10^k} = \frac{10^k - 9^k}{10^k} = 1- \left(\frac{9}{10}\right)^k.$$ Since the above limit goes to 1 as $k\rightarrow \infty$, almost all the numbers contain the digit 9, or any digit for that matter.