-- The highest 4-digit integer is 9999.
-- The lowest one is 1000.
-- So what you're really asking is:
"How many counting numbers are there past 999 until you get to 9999 ?"
Obviously, there are 9,000 of them.
22 of them.
4
from 1000 to 9999, ie 9999 - 999 = 9000
You're asking for every other number from 1,000 to 9,998 . There are 4,500 of them.
Four random, positive, one-digit integers.
22 of them.
4
18 positive integers and 36 integers (negative and positive)
from 1000 to 9999, ie 9999 - 999 = 9000
You're asking for every other number from 1,000 to 9,998 . There are 4,500 of them.
Four random, positive, one-digit integers.
The positive integers up to 4 are: 1, 2, 3, and 4. This is a total of four positive integers.
Fifty
-22
There are 4500 such numbers.
900 This explains it. A positive integer is a palindrome if it reads the same forward and backwards such as 1287821 and 4554. Determine the number of 5-digit positive integers which are NOT palindromes. We start by counting the total number of 5 digit positive integers. The first digit is between 1 and 9, so we have 9 choices. Each of the other 4 digits can be anything at all (10 choices for each). This gives us 9(10)4 = 90000 five-digit positive integers. Now we need to count the number of 5 digit palindromes. Again, we have 9 choices for the first digit and 10 choices for each of the next two. The tens and units digits however are fixed by our choices so far. Therefore, there are only 900 five-digit palindromes. Therefore, the total number of five-digit positive integers which are not palindromes is 90000-900 = 89100.
you need to do the same thing as 2 digit integers is really easy