any multpile of 18
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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.
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
Also, remember, that if a number is divisible by 9, then it is automatically divisible by three.
For each given number, add the digits. If the sum comes to nine, then it is divisible by '9'. 234 ; 2 + 3 + 4 = 9 So is dividble by '9' 345 ; 3 + 4 + 5 = 12 = 1 + 2 = 3 Not divisible by '9' 567 ; 5 + 6 + 7 = 18 = 1 + 8 = 9 so is divisible by '9'.
Here are two digit numbers that are divisible by both 6 and 9: 18 (which is the least common multiple) 36 54 72 90