Oh, isn't that a happy little question! Let's take a look at our number 2385. To see if it's divisible by another number, we can check if it can be evenly divided without any remainders. You can try dividing it by different numbers like 2, 3, 5, and so on, and see which ones give you a whole number answer. Just remember, there are no mistakes, just happy little discoveries!
2400
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.
999 is divisible by 9, but not by six; the next lower number divisible by 9 is 990, which is also divisible by 6, so that's the answer. Some shortcuts for divisibility: 0 is divisible by any number. If the last digit of a number is divisible by 2, the number itself is divisible by 2. If the sum of the digits of a number is divisible by 3, the number itself is divisible by 3. If the last TWO digits of a number are divisible by 4, the number itself is divisible by 4. If the last digit of a number is divisible by 5, the number itself is divisible by 5. If a number is divisible by both 2 and 3, it is divisible by 6. If the last THREE digits of a number are divisible by 8, the number itself is divisible by 8. If the sum of the digits of a number is divisible by 9, the number itself is divisible by 9. 990: 9+9+0=18, which is divisible by 9, so 990 is divisible by 9. 18 is also divisible by 3, so 990 is divisible by 3, and since 990 ends in 0 it's also divisible by 2, meaning that it's divisible by 6 as well.
35 is not divisible by 3.
403 is divisible by no number
yes
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)
Any of its factors
No. Numbers divisible by 10 end with a zero.
Yes, the result is 795.
1, 3, 5, 9, 15, 45, 53, 159, 265, 477, 795, 2385
Yes but not evenly, yields a remainder
The numbers that are divisible by 795 are infinite. Some of them are: 795, 1590, 2385, 3180 . . .
not without a remainder/decimal. the answer is 589 remainder 2 or 589.5
4773
2385
1, 3, 5, 9, 15, 27, 45, 53, 135, 159, 265, 477, 795, 1431, 2385, 7155.