$\frac{7x+1}2, \frac{7x+2}3, \frac{7x+3}4, \ldots ,\frac{7x+2016}{2017}$ are reduced fractions for integers $x\in(0,301)$. [closed]

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 $$4$$.

closed as off-topic by Namaste, zipirovich, mrtaurho, KReiser, Lee David Chung LinJan 9 at 10:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, zipirovich, mrtaurho, KReiser, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.

• What do you mean? Any rational number can be written in the way you want, no? – lulu Jan 8 at 14:36
• $7x+\dfrac12$ or$\dfrac{7x+1}2$ – lab bhattacharjee Jan 8 at 14:39
• $\dfrac{7x+1}2$ – Shromi Jan 8 at 14:44
• Your question isn't clear. Perhaps it would help to try an example. Is $x=1$ an example of what you want? Why or why not? – lulu Jan 8 at 14:45
• So, we need $$(7x-1,n)=1$$ sufficient to check for primes $\le2017$ – lab bhattacharjee Jan 8 at 14:47

As noted by lab bhattacharjee, we need to see if $$7x-1$$ is co-prime to all $$k=2,3,\ldots,2017$$. If $$0, then $$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.

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

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

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

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

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

6. $$x=300$$ so $$7x-1=2099$$ is prime so it is another good candidate.

Therefore there are only two good $$x$$s: $$x=290$$ and $$x=300$$.

Assume $$n, m$$ are positive integers such that $$n. Given a fraction $$\frac{m}{n}$$, where $$\gcd(m, n)=g, this fraction can be simplified to a a fraction with co-prime numerator and denominator by dividing both $$m$$ and $$n$$ by $$g$$, i.e., $$\gcd(\frac{m}{g}, \frac{n}{g})=1$$.

In our case, we need to find all $$x<301$$ such that for all $$i=1, 2, ..., 2016$$: $$\gcd(7x+i, i+1) < i+1$$ as $$i+1$$ is the smaller of the two.

Since $$\gcd(a, b)=\gcd(a-b,b)$$ we get $$\gcd(7x+i,i+1)=\gcd(7x-1,i+1) If $$i+1$$ is prime, this becomes $$\gcd(7x-1, i+1)=1$$ because $$1$$ is the only factor of a prime number less than the number itself. As long as the chosen number $$x$$ satisfies the above equality for all primes between $$2$$ and $$2017$$ then it will automatically satisfy it for all other numbers in the same range (as every other number will be a product of some of these primes).

Now notice that if $$7x-1\leq 2017$$ then there will always be a number, namely, $$i=7x-2<2017$$, that will make the above equality fail ($$\gcd(a,a)=a>1$$). So we need $$7x-1>2017$$, which means that $$x\geq 289$$. Also notice that if $$x$$ is odd, then the first fraction would simplify. So all is left to check are the numbers $$\{290, 292, 294, 296, 298, 300\}$$.

The only solutions are $$290$$ and $$300$$.

• Thanks for answer. But it hard for an 8th grade student to understand such an answer. Could you make it simple. – Shromi Jan 8 at 16:17
• Is there any part in particular where you got lost? – EuxhenH Jan 8 at 16:33