Here are two digit numbers that are divisible by both 6 and 9:
18 (which is the least common multiple)
36
54
72
90
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261, 264, or 267
6 + 4 + 6 = 16 1 + 6 = 7 → No; 646 is not divisible by 9 (there is a remainder of 7). ----------------------------------------- Only if the sum of the digits is divisible by 9 is the original number divisible by 9. Repeat the test on the sum until a single digit remains; only if this single digit is 9 is the original number divisible by 9, otherwise this single digit is the remainder when the original number is divided by 9.
42, 84
If the last digit of a number is 0, 2, 4, 6, 8 or 0 then the number is divisible by 2. The digit of 619 is 9, so 619 is not divisible by 2.
No. 189 is only evenly divisible by 3 and 9 (from the set provided). Using the following rules of divisibility on the number 189: Divisible by 2? No - the number is not even Divisible by 3? Yes - the sum of the digits (1 + 8 + 9 = 18) is divisible by 3 Divisible by 4? No - the last two digits are not evenly divisible by 4 Divisible by 5? No - the last digit is not a 0 or a 5 Divisible by 6? No - the number is not even Divisible by 9? Yes - the sum of the digits is divisible by 9 Divisible by 10? No - the number is not divisible by 2 or 5