To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8};
The last digit is 0, which is one of these so 330 is divisible by 2.
To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3.
As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3.
330→ 3 + 3 + 0 = 6
6 is one of {3, 6,9} so 330 is divisible by 3.
To be divisible by 5, the last digit must be one of {0, 5}.
The last digit is 0 which is one of {0, 5} so 330 is divisible by 5.
There is no real check for 7 which is not much slower than just dividing by 7 to see if there is no remainder. One check:
Write the digits in blocks of 3 starting from the right hand end (like you would for reading the number):
in each block of 3 add twice the first digit to three times the second digit to the third digit.
Alternately subtract and add the blocks starting from the right hand end of the number.
If the result is divisible by 7, then so is the original number.
330 → 2×3 + 3×3 + 0 = 15
15 is not divisible by 7, so 330 is not divisible by 7.
To be divisible by 11, alternately subtract and add the digits of the number from the right hand end; only if this sum is divisible by 11 (or is 0) is the original number divisible by 11.
330 → 0 - 3 + 3 = 0
0 is 0, so 330 is divisible by 11.
Therefore 330 is divisible by 2, 3, 5, 11
But not divisible by 7.
Any even number is divisible by 2.
Factors of numbers are divisible by them with no remainders
If the last 4 digits are divisible by 80, the entire number is divisible by 80.But really, it is hardly worth-while to learn divisibility rules for a large amount of numbers; only for a few small numbers. Normally it is easier to just do the division.
The answer will depend on the divisibility rules list.
If the number is even and the sum of its digits is divisible by nine then the number is divisible by 18.
The divisibility rules for a prime number is if it is ONLY divisible by 1, and itself.
Any even number is divisible by 2.
Factors of numbers are divisible by them with no remainders
If a number is divisible by 3, and also by 4, then it is divisible by 12 - so you might use the divisibility rules for those two numbers. Although it might be simpler just to perform the division.
If the last 4 digits are divisible by 80, the entire number is divisible by 80.But really, it is hardly worth-while to learn divisibility rules for a large amount of numbers; only for a few small numbers. Normally it is easier to just do the division.
If a number is divisible by 3 and 5, it is divisible by 15.
a number is divisible by 9 if the sum of the digits is divisible by 9.
Yes.
The answer will depend on the divisibility rules list.
Divisibility rules help you find the factors of a number. Once you've found the factors for two or more numbers, you can find what they have in common. Take 231 and 321. If you know the divisibility rules, you know that they are both divisible by 3, so 3 is a common factor.
With the common divisibility rules, you can quickly see that it is divisible by 5, and by 9 (3 x 3). If you divide 225 by each of these numbers, you should be able to get the remaining factors quickly, as well.
If the last 3 digits are divisible by 8 and the sum of the digits are divisible by 9.