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.
100 times greater.
With a total of 90,000 five digit numbers, we can conclude that there is 45,000 EVEN five digit numbers.
The answer depends on what the tens digit is greater than, and what the ones digit does then.
There are five such numbers.