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No. Every third consecutive natural number is divisible by 3.

Q: Can you find three consecutive natural numbers none of which is divisible by 3?

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The sum of three consecutive odd numbers must be divisible by 3. As 59 is not wholly divisible by 3 the question is invalid. PROOF : Let the numbers be n - 2, n and n + 2. Then the sum is 3n which is divisible by 3. If the question refers to three consecutive numbers then a similar proof shows that the sum of these three numbers is also divisible by 3. Again, the question would be invalid.

Any three consecutive integers are divisible by three because it can be shown that the sum divided by three is the middle number.

Yes, if the first number is odd.

There must be three consecutive integers to guarantee that the product will be divisible by 6. For the "Product of three consecutive integers..." see the Related Question below.

The answer to that question is 49,50,51

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0, 1, 2

That doesn't work. The number has to be divisible by three. Any three consecutive numbers add up to a multiple of three.

5+2+1=8 and 8 is not divisible by 3.

No.

The sum of three consecutive odd numbers must be divisible by 3. As 59 is not wholly divisible by 3 the question is invalid. PROOF : Let the numbers be n - 2, n and n + 2. Then the sum is 3n which is divisible by 3. If the question refers to three consecutive numbers then a similar proof shows that the sum of these three numbers is also divisible by 3. Again, the question would be invalid.

Any three consecutive integers are divisible by three because it can be shown that the sum divided by three is the middle number.

2, 3Those two are consecutive, natural and prime numbers! It's as easy as one, two, three! (Pun intended)

Yes, if the first number is odd.

It is a statement of numerical fact.

No, since 2 is not divisible by three in the natural numbers.

The sum of any three consecutive even numbers must be divisible by 3. 32 is not, so there is no solution.

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