The smallest one in that range is 1,000. The largest one in that range is 6,000.All the numbers between 1,000 and 6,000 are 4-digit numbers. There are6,000 - 999 = 5,001of them.
None. 3 digit numbers are not divisible by 19 digit numbers.
With 123 digits you can make 123 one-digit numbers.
None. The sum of one digit can't be twice the size of the digit.
I suggest you calculate (1) how many numbers are in total, and (2) how many numbers in that range do NOT have 6 as a digit. Then you can subtract the result of (1) minus the result of (2).
19
The smallest one in that range is 1,000. The largest one in that range is 6,000.All the numbers between 1,000 and 6,000 are 4-digit numbers. There are6,000 - 999 = 5,001of them.
None. 3 digit numbers are not divisible by 19 digit numbers.
i need it for homework too
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 9 possible numbers for the first digit (one of {1, 2, ..., 9}); with 9 possible digits for the second digit (one of {0, 1, 2, ..., 9} which is not the first digit)); with 8 possible digits for the third digit (one of {0, 1, 2, ..., 9} less the 2 digits already chosen); This there are 9 × 9 × 8 = 648 such numbers.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
Only one . . . 999 .
With 123 digits you can make 123 one-digit numbers.
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
None. The sum of one digit can't be twice the size of the digit.
2,4,6,8,10,12,14,16,18