What are the divisibility rules to determine which numbers 330 is divisible by from the numbers 2 3 5 7 11 and what are those numbers?

Answer 1

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.