# Another simple problem

arcnets
Prove this: Among any 6 natural numbers in a row (e.g. 20,21,22,23,24,25) there's at least 2 of them which have no common divisor larger than 1.

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.

arcnets
Good, Ben-CS. Except the L in LCD means largest not least. - Anyone?

We're both wrong: It should be GCF, for Greatest Common Factor. (Those *...* fractions are messing with my mind!)

call the numbers x, x+1, x+2...x+5

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.

Njorl

arcnets
Correct IMO, Njorl.

ottjes
Another problem:

consider a set of 11 numbers

Define the that sets can you do the following:
Remove 1 arbritary number from the set then divide the set into 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.