To be palindromic, the year must be of the form XYYX where X and Y are digits, so we need only consider the first half of the number, namely the XY and the question becomes how many numbers XY are there between 20 and 99 such that XYYX is between 2000 and 9999.
The answer depends upon the definition of "between":
XY = 99 ⇒ 9999 = 9999 so not allowed
⇒ number of years is 99 - 20 = 79
XY = 99 ⇒ 9999 = 9999 so still allowed
⇒ number of years is 99 - 20 + 1 = 80
The sum of all palindromic numbers from 1001 to 9999 is 495000.
11000
.9999 is 10 times purer (10 times less impurities).
9999
90 palindromes.
5445 is one that I can think of
That would be every number from 20 to 99 followed by the number backward (2002, 2112, 2222, 2332, ..., 9779, 9889, 9999). The quantity is 99 - 19 or 80.
The sum of all palindromic numbers from 1001 to 9999 is 495000.
The difference between 9999 and 999.9 is: 9999 - 999.9 = 8999.1
11000
a 8-digits palindromic number is a number consisting of a 4-digits number written and then written backward, i.e. 1234 4321 so there are as many palindromic 8-digits numbers as 4-digits numbers so from 1000 to 9999, so there are 8999 palindromic 8 digits number (I assumed that 00011000 is not to be considered as a valid 8 digit number)
If by "between" you mean "between but not including", then the answer is 9999 - 999 - 1 = 8999.If on the other hand, you mean to include both 9999 and 999, then the answer is 9999 - 999 + 1 = 9001.
his story was in the who cares era, in a summer of 9 99 999 9999 9999 9999 9999 years ago.
There are 2828 integers between 1000 and 9999.
Prime No: between 1000 to 9999= 54_( 42x3)
There are 10 palindromes divisible by 9 between 1000 and 9999.
.9999 is 10 times purer (10 times less impurities).