Show that any sequence of 10 consecutive positive integers contains at least one number that is relatively prime to the product of all the others.
The following lemma suffices:
Lemma : In any sequence of 10 consecutive positive integers, there is one that is divisible by neither 2, 3, 5 or 7.
Once we have the lemma, it suffices to pick an integer x with the property guaranteed by lemma. If x is not relatively prime to the product of all others, then it is not relatively prime to one of them, let's say y.
In particular, gcd(x,y) is more than 1 but, by Euclid's algorithm, gcd(x,y) divides |x-y| which is at most 9. It follows that one of 2,3,5 or 7 must divide gcd(x,y) and hence x, contradiction.
It remains to prove lemma. Note that it suffices to prove the lemma for integers up to 210=2*3*5*7.
There are different ways to prove this, I simply ran a small program, here are the integers up to 210 satisfying the conclusion of the lemma :
1
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97
101
103
107
109
113
121
127
131
137
139
143
149
151
157
163
167
169
173
179
181
187
191
193
197
199
209
There is no gap greater than 10 (in fact, only 2 gaps of size 10, at the beginning and the end).
