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Yes because 6+4+2 is divisible by three.

The simple way to decide if any number is entirely divisible by three is to first discard any digit in the number that is divisible by three (3, 6, 9) then add any of the digits remaining together and if at any point you get something that is divisible by three you can disregard those digits. If at the end you have a remainder, the number is not evenly divisible by three. If there is no remainder the entire number is divisible by three.

In your example 642, I threw away the 6 leaving 4 and 2 , added those = 6 which is divisible by three (actually I remembered that 42 was divisible by 6 so I knew right after I had taken away the first 6 that your number was divisible by 3.

Q: Is 642 divisible by 3

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No. 642 is not evenly divisible by nine.

You can add up the digits, 6+4+2=12 and if that number is divisible by three, so it the original number. So 12 is certainly divisible by 3 so that means 642 is also.

Yes, evenly

Yes and it is 107

5

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No. 642 is not evenly divisible by nine.

642 is composite. It is divisible by 2, 3, 107, 214, and 321.

You can add up the digits, 6+4+2=12 and if that number is divisible by three, so it the original number. So 12 is certainly divisible by 3 so that means 642 is also.

1, 2, 3, 6, 107, 214, 321, 642

Just by looking at it, without trying any factors to see if they work, you can see-- 642 is even, i.e. divisible by 2, since it ends in '2'-- 642 is divisible by 3, since its digits add up to (6+4+2) = 12 which is divisible by 3.Just looking at 642, you already know of at least 2 factors that it has besides '1' and itself,so it can't be a prime number.

Yes, evenly

Yes and it is 107

5

here are some 2,3,4,6,8,12,13,16,24,26,32,48,52,321&642

Only by 6 evenly with no remainder

642 x 3 = 1926

No, the first give-away is that it is an even number (divisible by 2).