Proving Co-Prime Numbers in Sets of Five

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In any set of five consecutive integers, at least one number is guaranteed to be co-prime to the other four. This conclusion is derived from the properties of prime numbers and their distribution among integers. For instance, in the set (2, 3, 4, 5, 6), the number 5 is co-prime to 2, 3, 4, and 6. This principle holds true for any selection of five consecutive numbers, confirming the existence of at least one co-prime integer within the set.

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kaliprasad
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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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