1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
(10,20,21 etc etc)
There is: 101,111,121,131,141,151,161,171,181,191 202,212,222,etc... 999 There are 90 palindromic 3 digit numbers
There is only one such number.
None. 3 digit numbers are not divisible by 19 digit numbers.
100 times greater.
There are 21 two-digit prime numbers.
How many two digit numbers are there in which the tens digit is greater than the oneβs digit ?
There are 45 of them.
Ten.
There is: 101,111,121,131,141,151,161,171,181,191 202,212,222,etc... 999 There are 90 palindromic 3 digit numbers
ccsndf
-3
There is only one such number.
None. 3 digit numbers are not divisible by 19 digit numbers.
The answer depends on what the tens digit is greater than, and what the ones digit does then.
With a total of 90,000 five digit numbers, we can conclude that there is 45,000 EVEN five digit numbers.
There are five such numbers.
There are 21 two-digit prime numbers.