Q: What is the largest 2-digit number divisible by 3?

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The number 100,002 is divisible by 3. The largest 5 digit number is 99996 (6 less).

99

Converse:If a number is divisible by 3, then every number of a digit is divisible by three. Inverse: If every digit of a number is not divisible by 3 then the number is not divisible by 3? Contrapositive:If a number is not divisible by 3, then every number of a digit is not divisible by three.

999 is divisible by 9, but not by six; the next lower number divisible by 9 is 990, which is also divisible by 6, so that's the answer. Some shortcuts for divisibility: 0 is divisible by any number. If the last digit of a number is divisible by 2, the number itself is divisible by 2. If the sum of the digits of a number is divisible by 3, the number itself is divisible by 3. If the last TWO digits of a number are divisible by 4, the number itself is divisible by 4. If the last digit of a number is divisible by 5, the number itself is divisible by 5. If a number is divisible by both 2 and 3, it is divisible by 6. If the last THREE digits of a number are divisible by 8, the number itself is divisible by 8. If the sum of the digits of a number is divisible by 9, the number itself is divisible by 9. 990: 9+9+0=18, which is divisible by 9, so 990 is divisible by 9. 18 is also divisible by 3, so 990 is divisible by 3, and since 990 ends in 0 it's also divisible by 2, meaning that it's divisible by 6 as well.

35 is not divisible by 3.

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The number 100,002 is divisible by 3. The largest 5 digit number is 99996 (6 less).

5 multiplication table always ends with 5 or 0 so that largest 3 digit no is 995

There is no such number. If you have any such number, n, that is divisible by 3 and 5 then n + 15 is larger, and is divisible by both. And you can add another 15 to that number, and then to that, for ever more.

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99

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