9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 135, 144, 153, 162, 171, 180, 189, 198, 207, 216, 225, 234, 243, 252, 261, 270, 279, 288, 297, 306, 315, 324, 333, 342, 351 and keep adding nine until you get to 999.
109 whole numbers greater than 9 and less than 999 are multiples of 9
3, 9, 15, 21, 27, 33 and just keep adding 6 until you get to 999.
The numbers 3,6 and 9 are all multiples of 3 because 3 can go into all of them evenly.
9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99 and just keep adding 9 until you get to 396.
There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for each of the first digits, There are 8 possible numbers {1-4, 6-9} for each of the first two digits, Making 9 x 10 x 8 = 720 possible 3 digit counting numbers not multiples of 5.
Way too many to list... 3 6 9 12 15 18 21 ...
109 whole numbers greater than 9 and less than 999 are multiples of 9
999
9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99 and just keep adding 9 until you get to 999.
All multiples of 90 are divisible by 9 and 10; thus all multiples of 90 between 100 and 999 inclusive solve the problem, that is any of: 180, 270, 360, 450, 540, 630, 720, 810, 900, 990.
They are: 9 18 27 36 45 54 and 63
Yes, all multiples of 9 are also multiples of 3.but they're not all of them. Every multiple of 9 is also a multiple of 3, but there are more multiples of 3 besides those.
Common multiples of 9 and 10 are all of the multiples of 90.
All multiples of 18 are divisible by 9. Not all multiples of 9 are divisible by 18.
There are infinitely many multiples of 9 and it is not possible to add them all.
yes since 9 is divisible by 3 all its multiples are as well
No; all multiples are not multiples of 6, for example 3×3 = 9 is not a multiple of 6. However as 6 = 2×3 all multiples of 6 are also multiples of 3.