Are two integers coprime if they are coprime mod(n)?

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Homework Statement


Show that out of a set of ten consecutive integers, at least one is coprime to all of the others.

Homework Equations


Lemma: Out of a set of n consecutive integers, exactly one is divisible by n. (Given).

The Attempt at a Solution


Let a1, a2...a10 be consecutive integers. Let a11(mod10), a2≡ 2(mod10)...a100(mod10).

Observe that 7 is coprime to the other integers from 1 to 10. Does this mean that 7(mod10) is coprime to the other integers in ℤ10, and if so, does this mean that a7 is coprime to the other ai?

Also, sorry that I've been throwing so many questions on here in the last few weeks. Thank you all for being so patient :)
 
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Never mind, 7 has a multiplicative inverse, 3, in the integers modulo 10, so it can divide the other elements.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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