<s>[INST] $\frac{7x+1}2, \frac{7x+2}3, \frac{7x+3}4, \ldots ,\frac{7x+2016}{2017}$ are reduced fractions for integers $x\in(0,301)$. 
BdMO 2017 junior catagory Question 7. $$\dfrac{7x+1}2, \dfrac{7x+2}3, \dfrac{7x+3}4, \ldots ,\dfrac{7x+2016}{2017}$$
  Here $x$ is a positive integer and $x < 301$. For some values of $x$ it is possible to express these given fraction in such fraction where denominator and numerator are co-prime. How many such $x$ is possible?

For an example if $x=4$, then $\dfrac{7x+1}2 = \dfrac{28+1}2 = \dfrac{29}2$. Here $29$ and $2$ are co-primes. But in the third term of this pattern I've noticed that $\dfrac{28+2}3 = \dfrac{30}3$ where $30$ and $3$ are not co-primes. So, $x$ is not $4$.
 [/INST] As noted by lab bhattacharjee, we need to see if $7x-1$ is co-prime to all $k=2,3,\ldots,2017$.  If $0<x<301$, then $6\le 7x-1\le 7\times 300-1=2099$.  Clearly, an integer in the interval $[6,2017]$ is not co-prime to all of the integers in $[2,2017]$ (in other words, $t$ is not co-prime to $t$ for $t\ge 2$).  So we have to worry about $x$s such that $2017<7x-1\le 2099$, that is, $289\le x\le 300$.  Plus if $x$ is odd, $7x-1$ is even so it is not co-prime to $2$.  There are only $6$ integers remaining, and the list of $7x-1$ is as shown below.


*

*$x=290$ so $7x-1=2029$ is prime so it is a good candidate.

*$x=292$ so $7x-1=2043$ is not co-prime to $3$.

*$x=294$ so $7x-1=2057$ is not co-prime to $11$.

*$x=296$ so $7x-1=2071$ is not co-prime to $19$.

*$x=298$ so $7x-1=2085$ is not co-prime to $5$.

*$x=300$ so $7x-1=2099$ is prime so it is another good candidate.
Therefore there are only two good $x$s: $x=290$ and $x=300$.
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