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As Richard observed in his posting, the trick with the number 6 is that there are no six consecutive numbers which are all divisible by 2, 3 and 5. Therefore one of the numbers must be divisible by a larger prime number p. (This works unless the numbers are 1, 2, 3, 4, 5, 6, in which case we can choose the prime number 5.) Since p<=7, p can divide only one of our 6 numbers.
The statement about the divisibility by 2, 3 and 5 can even be determined by merely trying out all possible combinations of remainders of 2, 3 and 5 on consecutive numbers, but below a sketchy "indirect" proof. So let's try to make all numbers divisible be 2, 3 and 5:
1. Only 3 numbers can be divisible by 2
2. Only 2 numbers can be divisible by 3. However one of these numbers must be even and was already counted as divisible by 2. Makes so far 4 numbers divisible by 2 or 3.
3. Now 2 numbers are missing. We can catch those if the first and last number are divisible by 5. But again, one of those two numbers was already counted as being divisible by 2. Makes only 5 numbers divisible by 2, 3 or 5.
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