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Easy if we use elimination (it helps to list down the numbers and cancel as we go along).
The number is even for obvious reason, so cancel 1 and 2.
If the number is not divisible by any odd n>1, then it is not divisible by 2n. but n and 2n is not beside each other, so we may cancel n and 2n for n=3,5,7,9,11,13,15.
By further elimination of number with no possible neighbor, we cancel 4,8,12,31.
Also,
divisible by 4,5 => divisible by 20
divisible by 3,8 => divisible by 24
divisible by 4,7 => divisible by 28.
Then we can cancel 20,24,28 which in turn cancels 19,21,23,25,27,29.
That leaves only the pair (16,17), which is the answer if we can verify the existence of a number which satisfies the condition.
However, this is straightforward, any number
8.7.11.13.17.19.23.25.27.29.31k
(where k is odd and not divisible by 17) works.
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