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, dude, let me break it down for you. Number 2385 is divisible by 1, 3, 5, 7, 15, 21, 35, 105, 227, 681, 1135, and 2385. So, if you ever need to impress someone at a math party, just drop that knowledge bomb on them. You're welcome.
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