There are infinitely many numbers divisible by 2, 3, 5 and 9. One such is 900.
30
2520
The answer is 120. 120/2= 60 120/3= 40 120/5= 24
3 and 5 are divisable by 195
its divisble by 2 9 3
No.
yes for a number to be divisble by 3 you have to add the numbers together and see if you can divide it by 3 evenly 1+5+3= 9 9 can be divided by 3 evenly so that makes the number, 153 divisble by 3
Factorization: 23 × 32 × 5 × 7 Ternary: 101101003 Binary: 1001110110002
yes it is because 90 has a 0 at the end if the end of the number is 0 or 5 it is divisble by 5
They are divisble by 5 because every number at the end if it ends in 0 or 5 then its divisble by 5.
by 3 . . . yes by 5 . . . no
yes 552 is divisible by 6 and the result is 92... to find out if a number is divisible by 6 then that number must be divisible by 2 and 3.. 1.If a number is divisble by three then the last digit must be divisible by 2 (so that the last digit must be 0,2,4,6,8) in this number the last digit is 2 so that it is divisible by 2 2.If a number is divisble by 3 then the sum of its digit must be divisible by 3. In this case the sum of the digits of 552 is 5+5+2= 12 and 12 is divisible by 3.... If this two rules fit in then that number is divisible by 6