at least one of 100 consecutive integers is relatively prime to all natural numbers less or equal 100 For an arbitrary integer $n$ define $A_n=\{i|n \leq i \leq n+99  \text{ where }i\text{ is an integer}\}$ (i.e. $A_n$ is 100 consecutive integers)  
Is it true that for any integer $n$ there is an element in $A_n$ which is relatively prime to all numbers in $A_1$?  
Thanks
 A: If we assume $0$ is relatively prime to all integers, perhaps we can use the Chinese remainder theorem to see more. Let $n\in\mathbb{Z}$. An element $m\in A_{n}$ will be relatively prime to all of $A_{1}$ iff it has nonzero remainder modulo every prime in $A_{1}$.
This gives a system of congruences
$$
m \equiv a_{1} (\bmod p_{1})\\
...\\
m \equiv a_{N} (\bmod p_{N})
$$ 
where $p_{1},...,p_{N}$ are the primes in $A_{1}$, and $a_{1},...a_{N}\neq 0$.
Let $M = p_{1}\cdots p_{N}$.
Then by the Chinese remainder theorem, $m = (a_{1}b_{1}\frac{M}{p_{1}} + ... + a_{N}b_{N}\frac{M}{p_{N}}) + tM$ for some $t\in\mathbb{Z}$. (here each $b_{i}$ is $a_{i}^{-1}$ taken in $\mathbb{Z}/p_{i}\mathbb{Z}$)
So I think a way to find the answer is to consider the set $B$ of smallest positive values of $a_{1}b_{1}\frac{M}{p_{1}} + ... + a_{N}b_{N}\frac{M}{p_{N}} + tM$ for all $a_{1},...a_{N}\neq 0$ and then make sure that $\forall b\in B$ $\exists c\in B$ with $|b-c|\leq 100$. If this is the case, then the answer is affirmative, if not, then the answer is negative (this is because $\forall b\in B$, $b+tM \not\equiv 0 (\bmod p_{i})$ $\forall i$, that is, $b+tM$ is relatively prime to all of $A_{1}$ $\forall t\in\mathbb{Z}$). For your definition of $A_{n}$, $|B| = (p_{1}-1)\cdots (p_{N}-1) = 277 399 690 427 737 839 953 078 806 118 400 000$ though...
We could do this for a much simpler scenario; suppose we define $A_{n} = \{i | n\leq i\leq n + 3\}$. The primes in $A_{1}$ are $2,3$, and the nonzero values of $a = (a_{1},a_{2})$ are $(1,1)$ and $(1,2)$, and $B = \{1,5\}$. Here $M = 2\cdot 3 = 6$. Since the distance between $1,5$ is $4$ and so not strictly greater than $4$, and since all translates of $1,5$ by multiples of $6$ will have nonzero remainder modulo all primes of $A_{1}$, all such translates will be relatively prime to every element of $A_{1}$. So for any $A_{n}$, there will be an element of $A_{n}$ relatively prime to all of $A_{1}$. 
