The first (hundreds) digit can be chosen in any one of 9 ways from the digits {1,2,3,...,9}.
With each one of these, the second (tens) digit can be chosen in any one of 10 ways from the digits {0,1,2,3,...,9}.
With each of these combinationsm the third (units) digit can be chosen in one of 5 ways from the digits {1,3,5,7,9}.
Altogether, then, there are 9*10*5 = 450 ways or 450 odd 3-digit numbers.
None. 3 digit numbers are not divisible by 19 digit numbers.
There are 21 two-digit prime numbers.
It would help to know which digit. 0 appears in 9 numbers and each of the others in 18 numbers.
There are 600 5-digit numbers divisible by 150.
17
4 Give examples to support your answer please.
The answer will depend on how many digits you have at your disposal. Working with the 10 digits, and barring leading zeros, there are 8,877,690 numbers.
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.
There are five such numbers.
There are 21 two-digit prime numbers.
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.
It would help to know which digit. 0 appears in 9 numbers and each of the others in 18 numbers.
The largest 3-digit number is 999The smallest 3-digit number is 100.There would be 999 of them, but the first 99 don't count. So there are 900 of them.
There are 600 5-digit numbers divisible by 150.
17
There are 17 such numbers.