1 digit number: only 1 number
2 digits number: 18 numbers
3 digits number: 76
So there are 95 numbers containing 9.
A nine-digit palindrome is structured as (a_1a_2a_3a_4a_5a_4a_3a_2a_1), where (a_1) must be between 1 and 9 (to ensure it's a nine-digit number), and (a_2), (a_3), and (a_4) can each be between 0 and 9. Therefore, there are 9 choices for (a_1), 10 choices each for (a_2), (a_3), and (a_4). This results in a total of (9 \times 10 \times 10 \times 10 = 9000) nine-digit palindromic numbers.
A five digit number with 9 is 99999. A number which is 1000 times larger (not 1000 times as large) is 99999000.
Nine of them.
The largest 9 digit number is 999,999,999 (nine hundred and ninety nine million, nine hundred and ninety nine thousand, nine hundred and ninety nine).
There are nine of them. They are:108117126135144153162171180
A nine-digit palindrome is structured as (a_1a_2a_3a_4a_5a_4a_3a_2a_1), where (a_1) must be between 1 and 9 (to ensure it's a nine-digit number), and (a_2), (a_3), and (a_4) can each be between 0 and 9. Therefore, there are 9 choices for (a_1), 10 choices each for (a_2), (a_3), and (a_4). This results in a total of (9 \times 10 \times 10 \times 10 = 9000) nine-digit palindromic numbers.
-8
There are no ten digit zip codes. Maryland has many five digit zip codes and many many more nine digit zip+4 codes. There can be a unique nine digit zip+4 code for every mailbox in a post office building and for at least every block in a town or city!
Nine million of them.
Yes. Twenty nine times.
A five digit number with 9 is 99999. A number which is 1000 times larger (not 1000 times as large) is 99999000.
900 It is the only three digit number with a nine in the numbers place
Nine of them.
How many four digit combinations can be made from the number nine? Example, 1+1+2+5=9.
The largest 9 digit number is 999,999,999 (nine hundred and ninety nine million, nine hundred and ninety nine thousand, nine hundred and ninety nine).
There are nine of them. They are:108117126135144153162171180
Instead of going from 1 to 500, let's just subtract one from the set and come up with "How many 3's appear from 1 to 499?" Total possible combinations in this, including leading zeros, is 5*10*10, i.e. 5 possible in the first set(0,1,2,3,4), and 10 in the second two(0,1,2,,4,5,6,7,8,9). Instead of figuring out how many have a 3, lets figure out how many don't. Well there is 4 possibilities in the first (0,1,2,4), and nine in the second two (0,1,2,4,5,6,7,8,9). So, total = 5 * 10 * 10 = 500 Without 3 = 4 * 9 * 9 = 324 500-324 = 176 There are 176 numbers from 1 to 500 that have the digit 3 in them.