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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)
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How many different two-digit numbers can be written if each digit must be one of the integers from 1 to 8 inclusive?
There are 8*8 = 64 such numbers.
How many 8 digit number are there in total?
The first 8 digit number is 10,000,000, the last is 99,999,999 which means there are 90,000,000 8 digit numbers
How many 10 digit palindromic numbers are there?
well, i think if you use this you can find out. A = 1-9 ,B = 0-9 , C = 0-9 , D = 0-9 , E = 0-9 for 2digit numbers = A A for 3 digit numbers = A B A for 4 digit numbers = A B B A and so on till you get to for 8 digit numbers = A B C D D C B A for 9 digit numbers = A B C D E D C B A and last for 10 digit number = A B C D E E D C B A this should work...
How many 4-digit numbers are there with no digit repeated?
9 x 9 x 8 x 7 = 4,536
How many page numbers have the digit 2 in the ones place and are divisible by 8?
13
