All you need too do in this case is prove that the LCD (Least Common Denominator) of two consecutive integers is 1. It then follows that m consecutive integers have at least m - 1 pairs of integers that have no common divisors greater than 1. But, since I am a Sadistic Mathematician, I shall leave the proof as an exercise for the reader.
Now, either x or x+1 is even, along with two other numbers. Regardless, we can redefine the set that is not divisible by 2 as y,y+2,y+4 with y defined as odd. The other numbers are multiples of 2.
Only one of these can be divisble by 3. Any multiple of 3 +/- 2 or 4 is not a multiple of 3.
We might think that if y is a multiple of 4, that y+4 would be as well, but none of them is even, so none of them is a multiple of 4.
For any higher multiples, if y=nz, where z>4, (n+1)z>y+4, ruling out higher multiples.
Define the that sets can you do the following:
Remove 1 arbritary number from the set then divide the set in to 2 subsets of 5 numbers each and make sure that the sum of the numbers of the first subset equals the sum of the numbers of the second subset.
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