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can you find three consective natural numbers none of which is divisible by 3?
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Nope
One of them has to be divisible by three because they are consecutive and there are three of them
All natural numbers can be written in the form\[3k+0\]\[3k+1\]\[3k+2\]If you were to increase a number by one you would get\[3k+3=3(k+1)=3q + 0\]which is of the form 3k+0 So no matter where you started, say for instance your first number was of the form 3k+1, our next number would be of the form 3k+2 and your third of the form 3k+0 -- divisible by three
1,2,3,4,5,6,7,8,9,10,11,... Every third number is divisible by three so you can't find three 'consecutive' numbers where none of them are divisible by three
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