No.
Not always because 33 is divisible by 3 but not by 9
150 ÷ 9 = 16 r 6 → the first natural number between 150 and 300 divisible by 9 is 9 × 17 (=154)300 ÷ 9 = 33 r 3 → the last natural number between 150 and 300 divisible by 9 is 9 × 33 (=297)→ there are 33 - 17 + 1 = 17 natural numbers between 150 and 300 divisible by 9.
No, is not divisible by 33. It is divisible by 32 so it can't be by 33.
99 is divisible by 1, 3, 9, 11, 33, and 99.
33
NO!!! To test if '9' divides into a number, then add the digits of the number. If they sum to '9' then it is divisible by '9' NB This test is only good for '9'. 33 = 3 + 3 = 6 So it doesn't add to nine , so '33' does not divide by '9' However. 36 = 3 + 6 = 9 . So 36 divides by '9' the answer is '4'.
1,089 is divisible by these numbers: 1, 3, 9, 11, 33, 99, 121, 363, 1089.
297 is divisible by (1 x 297) (99 x 3) (33 x 9) (27 x 11)
Correct, also applies to 9.yes, example: 99/3=33 9+9=18 18/3=6
22 and 33 can be eliminated because 22 is divisible by 2 and 33 is divisible by 3. We know from the multiplication that we learned that 3 x 9 = 27, so 29 is the prime number.
3, 9, 15, 21, 27, 33
99 is divisible by: 1, 3, 9, 11, 33, 99.