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 . . .
not without a remainder/decimal. the answer is 589 remainder 2 or 589.5
Yes, if x is an integer divisible by 3, then x^2 is also divisible by 3. This is because for any integer x, x^2 will also be divisible by 3 if x is divisible by 3. This can be proven using the property that the square of any integer divisible by 3 will also be divisible by 3.
yes
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.
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)
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
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!
1, 3, 5, 9, 15, 27, 45, 53, 135, 159, 265, 477, 795, 1431, 2385, 7155.
1, 3, 5, 9, 15, 27, 45, 53, 135, 159, 265, 477, 795, 1431, 2385, 7155
2385