Mixing
$frac{7x+1}2, frac{7x+2}3, frac{7x+3}4, ldots ,frac{7x+2016}{2017}$ are reduced fractions for integers...
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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 $4$.
elementary-number-theory contest-math fractions rational-numbers coprime
asked Jan 8 at 14:34
ShromiShromi
1098
$endgroup$
closed as off-topic by amWhy, zipirovich, mrtaurho, KReiser, Lee David Chung Lin Jan 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." – amWhy, 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.
|
show 1 more comment
$begingroup$
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 $4$.
elementary-number-theory contest-math fractions rational-numbers coprime
asked Jan 8 at 14:34
ShromiShromi
1098
$endgroup$
closed as off-topic by amWhy, zipirovich, mrtaurho, KReiser, Lee David Chung Lin Jan 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." – amWhy, 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.
$begingroup$
What do you mean? Any rational number can be written in the way you want, no?
$endgroup$
– lulu
Jan 8 at 14:36
$begingroup$
$7x+dfrac12$ or$dfrac{7x+1}2$
$endgroup$
– lab bhattacharjee
Jan 8 at 14:39
$begingroup$
$dfrac{7x+1}2$
$endgroup$
– Shromi
Jan 8 at 14:44
$begingroup$
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?
$endgroup$
– lulu
Jan 8 at 14:45
2
$begingroup$
So, we need $$(7x-1,n)=1$$ sufficient to check for primes $le2017$
$endgroup$
– lab bhattacharjee
Jan 8 at 14:47
|
show 1 more comment
$begingroup$
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 $4$.
elementary-number-theory contest-math fractions rational-numbers coprime
asked Jan 8 at 14:34
ShromiShromi
1098
$endgroup$
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 $4$.
elementary-number-theory contest-math fractions rational-numbers coprime
elementary-number-theory contest-math fractions rational-numbers coprime
asked Jan 8 at 14:34
ShromiShromi
1098
asked Jan 8 at 14:34
ShromiShromi
1098
edited Jan 8 at 16:23
user593746
asked Jan 8 at 14:34
ShromiShromi
1098
asked Jan 8 at 14:34
ShromiShromi
1098
asked Jan 8 at 14:34
ShromiShromi
1098
1098
closed as off-topic by amWhy, zipirovich, mrtaurho, KReiser, Lee David Chung Lin Jan 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." – amWhy, 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.
closed as off-topic by amWhy, zipirovich, mrtaurho, KReiser, Lee David Chung Lin Jan 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." – amWhy, 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.
$begingroup$
What do you mean? Any rational number can be written in the way you want, no?
$endgroup$
– lulu
Jan 8 at 14:36
$begingroup$
$7x+dfrac12$ or$dfrac{7x+1}2$
$endgroup$
– lab bhattacharjee
Jan 8 at 14:39
$begingroup$
$dfrac{7x+1}2$
$endgroup$
– Shromi
Jan 8 at 14:44
$begingroup$
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?
$endgroup$
– lulu
Jan 8 at 14:45
2
$begingroup$
So, we need $$(7x-1,n)=1$$ sufficient to check for primes $le2017$
$endgroup$
– lab bhattacharjee
Jan 8 at 14:47
|
show 1 more comment
$begingroup$
What do you mean? Any rational number can be written in the way you want, no?
$endgroup$
– lulu
Jan 8 at 14:36
$begingroup$
$7x+dfrac12$ or$dfrac{7x+1}2$
$endgroup$
– lab bhattacharjee
Jan 8 at 14:39
$begingroup$
$dfrac{7x+1}2$
$endgroup$
– Shromi
Jan 8 at 14:44
$begingroup$
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?
$endgroup$
– lulu
Jan 8 at 14:45
2
$begingroup$
So, we need $$(7x-1,n)=1$$ sufficient to check for primes $le2017$
$endgroup$
– lab bhattacharjee
Jan 8 at 14:47
$begingroup$
What do you mean? Any rational number can be written in the way you want, no?
$endgroup$
– lulu
Jan 8 at 14:36
$begingroup$
What do you mean? Any rational number can be written in the way you want, no?
$endgroup$
– lulu
Jan 8 at 14:36
$begingroup$
$7x+dfrac12$ or$dfrac{7x+1}2$
$endgroup$
– lab bhattacharjee
Jan 8 at 14:39
$begingroup$
$7x+dfrac12$ or$dfrac{7x+1}2$
$endgroup$
– lab bhattacharjee
Jan 8 at 14:39
$begingroup$
$dfrac{7x+1}2$
$endgroup$
– Shromi
Jan 8 at 14:44
$begingroup$
$dfrac{7x+1}2$
$endgroup$
– Shromi
Jan 8 at 14:44
$begingroup$
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?
$endgroup$
– lulu
Jan 8 at 14:45
$begingroup$
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?
$endgroup$
– lulu
Jan 8 at 14:45
2
2
$begingroup$
So, we need $$(7x-1,n)=1$$ sufficient to check for primes $le2017$
$endgroup$
– lab bhattacharjee
Jan 8 at 14:47
$begingroup$
So, we need $$(7x-1,n)=1$$ sufficient to check for primes $le2017$
$endgroup$
– lab bhattacharjee
Jan 8 at 14:47
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
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 $6le 7x-1le 7times 300-1=2099$. 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 $tge 2$). So we have to worry about $x$s such that $2017<7x-1le 2099$, that is, $289le xle 300$. 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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add a comment |
$begingroup$
Assume $n, m$ are positive integers such that $n<m$. Given a fraction $frac{m}{n}$, where $gcd(m, n)=g<n$, 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 $$gcd(7x+i,i+1)=gcd(7x-1,i+1)<i+1.$$ 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 $7x-1leq 2017$ 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 $xgeq 289$. 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 $300$.
answered Jan 8 at 15:21
EuxhenHEuxhenH
484210
$endgroup$
$begingroup$
Thanks for answer. But it hard for an 8th grade student to understand such an answer. Could you make it simple.
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– Shromi
Jan 8 at 16:17
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Is there any part in particular where you got lost?
$endgroup$
– EuxhenH
Jan 8 at 16:33
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $6le 7x-1le 7times 300-1=2099$. 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 $tge 2$). So we have to worry about $x$s such that $2017<7x-1le 2099$, that is, $289le xle 300$. 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 $x=300$.
$endgroup$
add a comment |
$begingroup$
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 $6le 7x-1le 7times 300-1=2099$. 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 $tge 2$). So we have to worry about $x$s such that $2017<7x-1le 2099$, that is, $289le xle 300$. 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 $x=300$.
$endgroup$
add a comment |
$begingroup$
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 $6le 7x-1le 7times 300-1=2099$. 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 $tge 2$). So we have to worry about $x$s such that $2017<7x-1le 2099$, that is, $289le xle 300$. 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 $x=300$.
$endgroup$
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 $6le 7x-1le 7times 300-1=2099$. 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 $tge 2$). So we have to worry about $x$s such that $2017<7x-1le 2099$, that is, $289le xle 300$. 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 $x=300$.
answered Jan 8 at 20:19
user614671
add a comment |
add a comment |
$begingroup$
Assume $n, m$ are positive integers such that $n<m$. Given a fraction $frac{m}{n}$, where $gcd(m, n)=g<n$, 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 $$gcd(7x+i,i+1)=gcd(7x-1,i+1)<i+1.$$ 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 $7x-1leq 2017$ 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 $xgeq 289$. 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 $300$.
answered Jan 8 at 15:21
EuxhenHEuxhenH
484210
$endgroup$
$begingroup$
Thanks for answer. But it hard for an 8th grade student to understand such an answer. Could you make it simple.
$endgroup$
– Shromi
Jan 8 at 16:17
$begingroup$
Is there any part in particular where you got lost?
$endgroup$
– EuxhenH
Jan 8 at 16:33
add a comment |
$begingroup$
Assume $n, m$ are positive integers such that $n<m$. Given a fraction $frac{m}{n}$, where $gcd(m, n)=g<n$, 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 $$gcd(7x+i,i+1)=gcd(7x-1,i+1)<i+1.$$ 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 $7x-1leq 2017$ 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 $xgeq 289$. 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 $300$.
answered Jan 8 at 15:21
EuxhenHEuxhenH
484210
$endgroup$
$begingroup$
Thanks for answer. But it hard for an 8th grade student to understand such an answer. Could you make it simple.
$endgroup$
– Shromi
Jan 8 at 16:17
$begingroup$
Is there any part in particular where you got lost?
$endgroup$
– EuxhenH
Jan 8 at 16:33
add a comment |
$begingroup$
Assume $n, m$ are positive integers such that $n<m$. Given a fraction $frac{m}{n}$, where $gcd(m, n)=g<n$, 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 $$gcd(7x+i,i+1)=gcd(7x-1,i+1)<i+1.$$ 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 $7x-1leq 2017$ 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 $xgeq 289$. 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 $300$.
answered Jan 8 at 15:21
EuxhenHEuxhenH
484210
$endgroup$
Assume $n, m$ are positive integers such that $n<m$. Given a fraction $frac{m}{n}$, where $gcd(m, n)=g<n$, 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 $$gcd(7x+i,i+1)=gcd(7x-1,i+1)<i+1.$$ 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 $7x-1leq 2017$ 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 $xgeq 289$. 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 $300$.
answered Jan 8 at 15:21
EuxhenHEuxhenH
484210
edited Jan 8 at 20:10
answered Jan 8 at 15:21
EuxhenHEuxhenH
484210
answered Jan 8 at 15:21
EuxhenHEuxhenH
484210
answered Jan 8 at 15:21
EuxhenHEuxhenH
484210
484210
$begingroup$
Thanks for answer. But it hard for an 8th grade student to understand such an answer. Could you make it simple.
$endgroup$
– Shromi
Jan 8 at 16:17
$begingroup$
Is there any part in particular where you got lost?
$endgroup$
– EuxhenH
Jan 8 at 16:33
add a comment |
$begingroup$
Thanks for answer. But it hard for an 8th grade student to understand such an answer. Could you make it simple.
$endgroup$
– Shromi
Jan 8 at 16:17
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Is there any part in particular where you got lost?
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– EuxhenH
Jan 8 at 16:33
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Thanks for answer. But it hard for an 8th grade student to understand such an answer. Could you make it simple.
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– Shromi
Jan 8 at 16:17
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Thanks for answer. But it hard for an 8th grade student to understand such an answer. Could you make it simple.
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– Shromi
Jan 8 at 16:17
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Is there any part in particular where you got lost?
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– EuxhenH
Jan 8 at 16:33
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Is there any part in particular where you got lost?
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– EuxhenH
Jan 8 at 16:33
add a comment |
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What do you mean? Any rational number can be written in the way you want, no?
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– lulu
Jan 8 at 14:36
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$7x+dfrac12$ or$dfrac{7x+1}2$
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– lab bhattacharjee
Jan 8 at 14:39
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$dfrac{7x+1}2$
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– Shromi
Jan 8 at 14:44
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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?
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– lulu
Jan 8 at 14:45
2
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So, we need $$(7x-1,n)=1$$ sufficient to check for primes $le2017$
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– lab bhattacharjee
Jan 8 at 14:47