From 1000 to 9999 (inclusive) there are 9999 - 1000 + 1 = 9000 integers. Half of which are even, half of which are odd. So the answer is 9000/2 = 4,500.
500 of them.
9*9*8*7 = 4536
Well, honey, the even integers between 1000 and 9999 can be found by dividing the total range by 2, since every other number is even. So, 9999 - 1000 = 8999, and 8999 divided by 2 gives you 4499 even integers. But remember, I'm not your calculator, so don't come crying to me if you can't count 'em all yourself!
One less than 10000.
9000 integers.
There are 2828 integers between 1000 and 9999.
from 1000 to 9999, ie 9999 - 999 = 9000
90 palindromes.
500 of them.
There are 3168 such numbers.
There are 120 of them.
9*9*8*7 = 4536
There are 1,000 positive integers between 1,000 and 9,999, inclusive, that are divisible by nine.
952 of them.
-- 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.
An infinite amount
There are 10 palindromes divisible by 9 between 1000 and 9999.