Oh, what a lovely question! If we're looking for 2-digit numbers where the one's digit is greater than the ten's digit, we simply need to think about the possibilities. There are 36 such numbers, ranging from 12 to 98. Just imagine all the happy little numbers waiting to be discovered!
To determine the number of 2-digit numbers where the units digit is greater than the tens digit, we need to consider the possible combinations. The tens digit can range from 1 to 9, while the units digit can range from 0 to 9. If the units digit is greater than the tens digit, there are 36 possible combinations (9 choices for the tens digit and 4 choices for the units digit).
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...
ccsndf
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
There are 84 such numbers.
There are 151 3-digit numbers that are divisible by 6.
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
There is no smallest whole number - negative numbers go on forever. Therefore, there are infinitely many whole numbers that are smaller than the greatest 2-digit number.
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...
ccsndf
-3
There is only one such number.
None. 3 digit numbers are not divisible by 19 digit numbers.
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.
100 times greater.