yes
1, 3, 5, 9, 15, 45, 53, 159, 265, 477, 795, 2385
Yes, the result is 795.
Any number which ends in 0 or 5 is divisible by 5. So, obviously 2385 is divisible by 5. This goes for base ten, not base 9 (I think)
1, 3, 5, 9, 15, 45, 53, 159, 265, 477, 795, 2385
Any of its factors
No. Numbers divisible by 10 end with a zero.
Yes but not evenly, yields a remainder
The numbers that are divisible by 795 are infinite. Some of them are: 795, 1590, 2385, 3180 . . .
Converse:If a number is divisible by 3, then every number of a digit is divisible by three. Inverse: If every digit of a number is not divisible by 3 then the number is not divisible by 3? Contrapositive:If a number is not divisible by 3, then every number of a digit is not divisible by three.
35 is not divisible by 3.
not without a remainder/decimal. the answer is 589 remainder 2 or 589.5
If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.