# How many positive integers between 0 and 1,000,000 contain the digit 9?

In the example, it says that its easier to count the number of integers that do not contain the digit 9

what i did is, since its from $0$ to $1{,}000{,}000$ i can simply say its to $999{,}999$ and for 6 digits there are 8 choices of numbers, so $6^8$. now subtract that from $1{,}000{,}000$ and we get $679{,}616$ as the numbers that contain the digit $9$, is that correct?

• (1) there are ten different digits (2) choosing 6 digits independently from a set of 8 would give you $8^6$ choices. – Joffan Apr 10 '17 at 20:54
• @Joffan why a set $8$ digits? There are $9$ digits that are not $9$, but you need to take into account whether leading zeroes are permitted. – Χpẘ Apr 10 '17 at 21:07
• It is easier to allow leading zeros and it doesn't affect the answer. 000xxx contains a 9 if and only if xxx contains a 9. – badjohn Apr 10 '17 at 21:09
• @user2460798 I am commenting on the question as written. You'll note I already mentioned the digit count. – Joffan Apr 10 '17 at 21:11
• @badjohn Right. Because a blank (or absence of a zero) is the same as a zero. Would've been different if the question was about the digit 0. – Χpẘ Apr 10 '17 at 21:16

You have six slots. In each slot you can place any digit between 0 and 9 (ten possibilities). This generates all $10^6$ possible numbers from 0 to $999,999$ (ignoring leading zeros as necessary). To generate a number that doesn't contain a nine, there are only nine possibilities for each slot (namely, $0$ through $8$), yielding $9^6$. So the total number of integers between zero and $999,999$ that do contain the digit nine is $10^6-9^6$. This is the same as the number of positive integers between 0 and a million that do so.

So while the "excluding leading zeros" thing can work and all, I find that allowing them actually simplifies the problem.

You have a list of 9 digits $\{0, 1, 2, ..., 8\}$ and you want to write all possible numbers with them of 6 characters (we'll add +1 for the number 1,000,000)

you get $9$ digits in $6$ different positions, leading to $9^6$ distinct numbers (these include those with leading 0s: 000,000; 000,010; 001,305; etc). So there's no need to worry about the 'length' of a number because we can just assume those of length 5 have one starting 0s, those with 4 have 2 starting 0s, etc. Otherwise you have to account for the position of the first non-zero digit and then add the possibility of 0's elsewhere for numbers like $050,000 = 50,000$

So up to length 6 (aka 888,888 [because anything past that will contain a 9 ]) we have $9^6$ distinct numbers and we just need to add 1 more to account for the number $1,000,000$ and we end up with $9^6+1$ total numbers between 0 and 1M that do not contain the digit 9

(if you want those that do, take $1,000,001-9^6-1$)

• The question does ask for 'positive integers', which 0 is not a part of, so you may want to handle that on your own. By definition 0 is NOT positive, but it was included in the range 0-1M provided, so it is possible they meant to exclude the endpoints on the range. – jorgeegroj Apr 10 '17 at 21:21

No, you forgot about $5$-digit, $4$-digit, $3$-digit, $2$-digit, $1$-digit numbers. Proceed in the same way as you have done

( As mentioned by @Joffan, it must be $8^6$ rather than $6^8$, not that too it should be 8.9^5) )

You will get :

$$999,999 - (8.9^5+8.9^4+8.9^3+8.9^2+8.9^1+8)= 999,999- \Big(8 \cdot \frac{9^6-1}{9-1}\Big) = 468,559$$

$8$ beacuse we have excluded $0$ in the first place.

First, using base 10, we find that there are 1,000,000 numbers between 0 and 1,000,000 that use the digits 0-9, and using base 9, converting 1,000,000 base 9 to base 10, we find that 531,441 numbers use numbers 0-8, subtracting the # of numbers that use 0-8 from the amount of numbers that use 0-9, we get that the amount of positive integers between 0 and 1,000,000 that use 9 is 468,559 numbers