MHB Proving Co-Prime Numbers in Sets of Five

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show that in a set of any 5 consecutive numbers there is at least one number that is co-prime to all the rest 4 (for example (2,3,4,5,6- 5 is co-prime to 2,3,4,6)
 
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Reducing modulo $5$, we see that it's enough to consider $\{1, 2, 3, 4, 5\}$ as any $5$ consecutive integers form a complete residue system. Since the observation is true for $\{1, 2, 3, 4, 5\}$, it is true for all case. QED.
 
mathbalarka said:
Reducing modulo $5$, we see that it's enough to consider $\{1, 2, 3, 4, 5\}$ as any $5$ consecutive integers form a complete residue system. Since the observation is true for $\{1, 2, 3, 4, 5\}$, it is true for all case. QED.

I am not convinced about the solution can you clarify it
 
because we have 5 consecutive number the largest difference is 4. so if there is a common factor between 2 numbers of the 5 it has to be <=4. So a common prime factor has to be 2 or 3.

now in a set of 5 consecutive numbers one of the numbers has to be of the form 6n + 1 or 6n - 1 which is neither divisible by 2 nor 3. so it is co-prime to rest of the 4.
 
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