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Q: Is 293 exactly divisible by 3?

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293 is not divisible by 5. Only numbers that end on 0 or 5 are exactly divisible by 5.No, it is not.

No. 293 is not evenly divisible by 20.

It is not divisible by 6. Note:If a number id divisible by 6 then it must be divisible by both 2 and 3.The above number is not divisible by 2 and 3 either.

It divides exactly by 3, but not 4.

No

It is not. 621 is divisible by 3, not the other way round.

There is no factorization of 293 because 293 is a prime number. It is divisible only by 1 and itself.

No because it will have a remainder

Yes and it is exactly 2129

Not exactly because it will have a remainder

not exactly...39.666 times

Not exactly because it will have a remainder

Yes it is. It goes in 206 times exactly.

20 is divisible by 2, 5 and 10. It is not exactly divisible by 3 or 9.

They are multiples of 3

Yes indeed. That's exactly what "divisible by 3" means.

No, 586 is only divisible by: 1, 2, 293, 586.

Yes, it is divisible by 2, 3 and 9, but is not exactly divisible by 5 or 10.

Not exactly because it will have remainder of 1

Not exactly because it will have remainder of 1

Not exactly because it will have remainder of 3

Not exactly because it will have a remainder of 1

293 is prime. It is only evenly divisible by itself and one.

No - because the sum of the digits does not divide exactly by 3.

No. 20/3 = 6.66666....with the 6 repeating.